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#### Page No 6.15:

#### Question 1:

Find the distance between the following pair of points:

(a) (−6, 7) and (−1, −5)

(b) (*a*+*b*, *b*+*c*) and (*a*−*b*, *c*−*b*)

(c) (*a*sinα, −*b*cosα) and (−*a*cos α, *b*sin α)

(d) (*a*, 0) and (0, *b*)

#### Answer:

The distance *d* between two points and is given by the formula

(i) The two given points are (−6, 7) and (−1, −5)

The distance between these two points is

Hence the distance is.

(ii) The two given points are

The distance between these two points is

Hence the distance is.

(iii) The two given points are and

The distance between these two points is

Hence the distance is.

(iv) The two given points are (*a, *0) and (0*, b*)

The distance between these two points is

Hence the distance is.

#### Page No 6.15:

#### Question 2:

Find the value of *a* when the distance between the points (3, *a*) and (4, 1) is $\sqrt{10}$.

#### Answer:

The distance *d* between two points and is given by the formula

The distance between two points (3*, a*) and (4*, *1) is given as. Substituting these values in the formula for distance between two points we have,

Now, squaring the above equation on both sides of the equals sign

Thus we arrive at a quadratic equation. Let us solve this now,

The roots of the above quadratic equation are thus 4 and −2.

Thus the value of ‘*a*’ could either be.

#### Page No 6.15:

#### Question 3:

If the points (2, 1) and (1, −2) are equidistant from the point (*x*, *y*), show that *x* + 3*y* = 0.

#### Answer:

The distance *d* between two points and is given by the formula

Here it is said that the points (2*, *1) and (1*, **−*2) are equidistant from (*x, y*).

Let be the distance between (2*, *1)* *and (*x*,* y*).

Let be the distance between (1, −2)* *and (*x, y*).

So, using the distance formula for both these pairs of points we have

Now since both these distances are given to be the same, let us equate both and

Squaring on both sides we have,

Hence we have proved that when the points (2*, *1) and (1*,−*2) are equidistant from (*x, y*) we have .

#### Page No 6.15:

#### Question 4:

Find the values of *x*,* y* if the distances of the point (*x*, *y*) from (−3, 0) as well as from (3, 0) are 4.

#### Answer:

The distance *d* between two points and is given by the formula

It is said that (*x, y*) is equidistant from both (−3,0) and (3,0).

Let be the distance between (*x, y*) and (−3,0).

Let be the distance between (*x, y*) and (3,0).

So, using the distance formula for both these pairs of points we have

Now since both these distances are given to be the same, let us equate both and .

Squaring on both sides we have,

It is also said that the value of both and is 4 units.

Substituting the value of ‘*x*’ in the equation for either or we can get the value of ‘*y*’.

Squaring on both sides of the equation we have,

Hence the values of ‘*x*’ and ‘*y*’ are.

#### Page No 6.15:

#### Question 5:

The length of a line segment is of 10 units and the coordinates of one end-point are (2, −3). If the abscissa of the other end is 10, find the ordinate of the other end.

#### Answer:

The distance *d* between two points and is given by the formula

Here it is given that one end of a line segment has co−ordinates (2*,−*3). The abscissa of the other end of the line segment is given to be 10. Let the ordinate of this point be ‘*y*’.

So, the co−ordinates of the other end of the line segment is (10*, y*).

The distance between these two points is given to be 10 units.

Substituting these values in the formula for distance between two points we have,

Squaring on both sides of the equation we have,

We have a quadratic equation for ‘*y*’. Solving for the roots of this equation we have,

The roots of the above equation are ‘*−*9’ and ‘3’

Thus the ordinates of the other end of the line segment could be.

#### Page No 6.15:

#### Question 6:

Show that the points (−4, −1), (−2, −4) (4, 0) and (2, 3) are the vertices points of a rectangle.

#### Answer:

The distance *d* between two points and is given by the formula

In a rectangle, the opposite sides are equal in length. The diagonals of a rectangle are also equal in length.

Here the four points are *A*(*−*4*,**−*1)*, B*(*−*2*,**−*4), *C*(4*,*0) and *D*(2*,*3).

First let us check the length of the opposite sides of the quadrilateral that is formed by these points.

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

The other pair of opposite sides are also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

For a parallelogram to be a rectangle we need to check if the diagonals are also equal in length.

Now since the diagonals are also equal we can say that the parallelogram is definitely a rectangle.

Hence we have proved that the quadrilateral formed by the four given points is a.

#### Page No 6.15:

#### Question 7:

Show that the points *A* (1, −2), *B* (3, 6), *C* (5, 10) and *D* (3, 2) are the vertices of a parallelogram.

#### Answer:

The distance *d* between two points and is given by the formula

In a parallelogram the opposite sides are equal in length.

Here the four points are *A*(1*,** −*2)*, B*(3*,* 6), *C*(5*,* 10) and *D*(3*, *2).

Let us check the length of the opposite sides of the quadrilateral that is formed by these points.

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

The other pair of opposite sides is also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

Hence we have proved that the quadrilateral formed by the given four points is a .

#### Page No 6.15:

#### Question 8:

Prove that the points *A*(1, 7), *B* (4, 2), *C*(−1, −1) *D* (−4, 4) are the vertices of a square.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are equal in length. Also, the diagonals are equal in length in a square.

Here the four points are *A*(1*, *7)*, B*(4*,* 2), *C*(*−*1*,** −*1) and *D*(*−*4*,* 4).

First let us check if all the four sides are equal.

Since all the sides of the quadrilateral are the same it is a rhombus.

For the rhombus to be a square the diagonals also have to be equal to each other.

Since the diagonals of the rhombus are also equal to each other the rhombus is a square.

Hence the quadrilateral formed by the given points is a.

#### Page No 6.15:

#### Question 9:

Prove that the points (3, 0), (6, 4) and (−1, 3) are vertices of a right-angled isosceles triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In an isosceles triangle there are two sides which are equal in length.

Here the three points are *A*(3*,* 0)*, B*(6*, *4) and *C*(*−*1*, *3).

Let us check the length of the three sides of the triangle.

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

We can also observe that

Hence proved that the triangle formed by the three given points is an.

#### Page No 6.15:

#### Question 10:

Prove that (2, −2) (−2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

#### Answer:

The distance *d* between two points and is given by the formula

In a right-angled triangle, by Pythagoras theorem, the square of the longest side is equal to the sum of squares of the other two sides in the triangle.

Here the three points are *A*(2*,**−*2)*, B*(*−*2*,*1) and *C*(5*,*2).

Let us find out the lengths of all the sides of the triangle.

Here we have,

Since the square of the longest side is equal to the sum of squares of the other two sides the given triangle is a .

In a right angled triangle the area of the triangle ‘*A*’ is given by,

In a right angled triangle the sides containing the right angle will not be the longest side.

Hence the area of the given right angled triangle is,

Hence the area of the given right-angled triangle is.

In a right-angled triangle the hypotenuse will be the longest side. Here the longest side is ‘*BC*’.

Hence the hypotenuse of the given right-angled triangle is

#### Page No 6.15:

#### Question 11:

Prove that the points (2*a*, 4*a*), (2*a*, 6*a*) and (2*a*+$\sqrt{3}a$, 5*a*) are the vertices of an equilateral triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In an equilateral triangle all the sides have equal length.

Here the three points are, and.

Let us now find out the lengths of all the three sides of the given triangle.

Since all the three sides have equal lengths the triangle has to be an equilateral triangle.

#### Page No 6.15:

#### Question 12:

Prove that the points (2,3), (−4, −6) and (1, 3/2) do not form a triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In any triangle the sum of lengths of any two sides need to be greater than the third side.

Here the three points are, and

Let us now find out the lengths of all the three sides of the given triangle.

Here we see that,

This is in violation of the basic property of any triangle to exist. Therefore these points cannot give rise to a triangle.

Hence we have proved that the given three points do not form a triangle.

#### Page No 6.15:

#### Question 13:

The points *A*(2, 9), *B*(*a*, 5) and *C*(5, 5) are the vertices of a triangle *ABC *right angled at *B*. Find the values of *a* and hence the area of Δ*ABC*.

#### Answer:

Consider the figure.

Using distance formula,

$\mathrm{AC}=\sqrt{{\left(2-5\right)}^{2}+{\left(9-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{{\left(-3\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{25}\phantom{\rule{0ex}{0ex}}{\mathrm{AC}}^{2}=25\mathrm{units}$

$\mathrm{AB}=\sqrt{{\left(2-a\right)}^{2}+{\left(9-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{AB}=\sqrt{\left(4+{a}^{2}-4a\right)+\left(16\right)}\phantom{\rule{0ex}{0ex}}\mathrm{AB}=\sqrt{4+{a}^{2}-4a+16}=\sqrt{{a}^{2}-4a+20}\phantom{\rule{0ex}{0ex}}{\mathrm{AB}}^{2}=({a}^{2}-4a+20)\mathrm{units}$

$\mathrm{BC}=\sqrt{{\left(5-a\right)}^{2}+{\left(5-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{BC}=\sqrt{\left(25+{a}^{2}-10a\right)+0}\phantom{\rule{0ex}{0ex}}\mathrm{BC}=\sqrt{{a}^{2}-10a+25}\phantom{\rule{0ex}{0ex}}{\mathrm{BC}}^{2}=({a}^{2}-10a+25)\mathrm{units}$

We are given that ABC is a right triangle right angled at B.

By Pythagoras theorem, we have;

${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25=({a}^{2}-4a+20)+({a}^{2}-10a+25)\phantom{\rule{0ex}{0ex}}\Rightarrow 25=2{a}^{2}-14a+45\phantom{\rule{0ex}{0ex}}\Rightarrow 2{a}^{2}-14a+20=0\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}-7a+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}-5a-2a+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow a(a-5)-2(a-5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (a-2)(a-5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=2ora=5$

We cannot put *a* = 5 as it will make BC = 0. So, we ignore *a* = 5 and accept *a* = 2.

$\Rightarrow {\mathrm{AB}}^{2}=4-8+20=16\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=4\mathrm{units}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{BC}}^{2}=4+25-20=9\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=3\mathrm{units}\phantom{\rule{0ex}{0ex}}\mathrm{And},\mathrm{AC}=5\mathrm{units}\phantom{\rule{0ex}{0ex}}\mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}\mathrm{Area}=\frac{1}{2}\times 3\times 4=6\mathrm{square}\mathrm{units}$

#### Page No 6.15:

#### Question 14:

Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3) and (−3,−2) is a rhombus.

#### Answer:

The distance *d* between two points and is given by the formula

In a rhombus all the sides are equal in length.

Here the four points are *A *(2*, **−*1)*, B *(3*, *4), *C *(*−*2*, *3) and *D *(*−*3*, **−*2).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

#### Page No 6.15:

#### Question 15:

Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.

#### Answer:

The distance *d* between two points and is given by the formula

In an isosceles triangle two sides will be of equal length.

Here two vertices of the triangle is given as *A *(2*, *0) and *B *(2*, *5). Let the third side of the triangle be *C*(*x, y*)

It is given that the length of the equal sides is 3 units.

Let us now find the length of the side in which both the vertices are known.

So, now we know that the side ‘*AB*’ is not one of the equal sides of the isosceles triangle.

So, we have

Equating these two equations we have,

Squaring on both sides of the equation we have,

We know that the length of the equal sides is 3 units. So substituting the value of ‘*y*’ in equation for either ‘*AC*’ or ‘*BC*’ we can get the value of ‘*x*’.

Squaring on both sides,

We have a quadratic equation for ‘*x*’. Solving for roots of the above equation we have,

Hence the possible co−ordinates of the third vertex of the isosceles triangle are.

#### Page No 6.16:

#### Question 16:

Which point on x-axis is equidistant from (5, 9) and (−4, 6) ?

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x*−axis which is equidistant from both the points *A *(5*, *9) and *B *(*−*4*, *6)

Let this point be denoted as *C*(*x, y*)

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 6.16:

#### Question 17:

Prove that the points (−2, 5), (0, 1) and (2, −3) are collinear.

#### Answer:

The distance *d* between two points and is given by the formula

For three points to be collinear the sum of distances between two pairs of points should be equal to the third pair of points.

The given points are *A *(*−*2*, *5), *B *(0*, *1) and *C *(2*, **−*3)

Let us find the distances between the possible pairs of points.

We see that

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence we have proved that the three given points are.

#### Page No 6.16:

#### Question 18:

The coordinates of the point *P* are (−3, 2). Find the coordinates of the point *Q* which lies on the line joining *P* and origin such that OP = OQ.

#### Answer:

If and are given as two points, then the co-ordinates of the midpoint of the line joining these two points is given as

It is given that the point ‘*P*’ has co-ordinates (*−*3*, *2)

Here we are asked to find out the co-ordinates of point ‘*Q*’ which lies along the line joining the origin and point ‘*P*’. Thus we can see that the points ‘*P*’, ‘*Q*’ and the origin are collinear.

Let the point ‘*Q*’ be represented by the point (*x, y*)

Further it is given that the

This implies that the origin is the midpoint of the line joining the points ‘*P*’ and ‘*Q*’.

So we have that

Substituting the values in the earlier mentioned formula we get,

Equating individually we have, and.

Thus the co−ordinates of the point ‘*Q*’ is

#### Page No 6.16:

#### Question 19:

Which point on *y*-axis is equidistant from (2, 3) and (−4, 1)?

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *y*-axis which is equidistant from both the points *A *(2*, *3) and *B *(*−*4*, *1).

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *y*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *y*-axis which lies at equal distances from the mentioned points is.

#### Page No 6.16:

#### Question 20:

The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.

#### Answer:

We are given three vertices of a parallelogram A(3, 4), B(3, 8) and C(9, 8).

We know that diagonals of a parallelogram bisect each other. Let the fourth vertex be D(*x, y*).

Mid point of BD = $\left(\frac{x+3}{2},\frac{y+8}{2}\right)$

Mid point of AC = $\left(\frac{9+3}{2},\frac{4+8}{2}\right)$ = $\left(6,6\right)$

Since the mid point of BD = mid point of AC

So, $\left(\frac{x+3}{2},\frac{y+8}{2}\right)$ = (6, 6)

Thus, *x* = 9, *y* = 4.

So, the fourth vertex is (9, 4).

#### Page No 6.16:

#### Question 21:

Find a point which is equidistant from the points $A\left(-5,4\right)andB(-1,6).$How many such points are there ?

#### Answer:

Let A(*x*, *y*) be the point which is equidistant from the points P(−5, 4) and Q(−1, 6).

Then AP = AQ.

$\Rightarrow A{P}^{2}=A{Q}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left[x-\left(-5\right)\right]}^{2}+{\left[y-4\right]}^{2}={\left[x-\left(-1\right)\right]}^{2}+{\left[y-6\right]}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+25+10x+{y}^{2}+16-8y={x}^{2}+1+2x+{y}^{2}+36-12y\phantom{\rule{0ex}{0ex}}\Rightarrow 10x-2x-8y+12y+41-37=0\phantom{\rule{0ex}{0ex}}\Rightarrow 8x+4y+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+y+1=0.....\left(i\right)$

Hence all the points satisfying the equation (i) are equidistant from the points P and Q.

There are infinite such points.

#### Page No 6.16:

#### Question 22:

The centre of a circle is ($2a,a-7)$. Find the values of *a* if the circle passes through the point (11,$-$9) and has diameter $10\sqrt{2}$ units.

#### Answer:

The length of the diameter is $10\sqrt{2}units$.

So, the radius is $5\sqrt{2}units$.

The centre of the circle be C(2*a*, *a*−7).

Suppose it passes through the point P(11, −9).

Therefore, PC = *r*

*$\Rightarrow P{C}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(11-2a\right)}^{2}+{\left(-9-a+7\right)}^{2}={\left(5\sqrt{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 121+4{a}^{2}-44a+{a}^{2}+4+4a=50\phantom{\rule{0ex}{0ex}}\Rightarrow 5{a}^{2}-40a+75=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(a-3\right)\left(a-5\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=3ora=5$*

Hence the values of *a* are 3 or 5.

#### Page No 6.16:

#### Question 23:

Ayush starts walking from his house to office . Instead of going to the office directly , he goes to a bank first , from there to his daughter 's school and then reaches the office. what is the extra distance travelled by Ayush in reaching the office ? ( Assume that all distances covered are in straight lines) . If the house is situated at (2,4) , bank at (5,8) , school at (13,14) and office at ( 13,26) and coordinates are in kilometers .

#### Answer:

The position of the ayush's house is (2, 4) and that of the bank is (5, 8).

The distance between the house and the bank is

${d}_{1}=\sqrt{{\left(5-2\right)}^{2}+{\left(8-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+16}\phantom{\rule{0ex}{0ex}}=5units$

The position of the the bank is (5, 8) and that of the school is (13, 14).

The distance between the bank and the school is

${d}_{2}=\sqrt{{\left(13-5\right)}^{2}+{\left(14-8\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{8}^{2}+{6}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{64+36}\phantom{\rule{0ex}{0ex}}=10units$

The position of the school is (13, 14) and that of the office is (13, 6)

The distance between the bank and the school is

${d}_{3}=\sqrt{{\left(13-13\right)}^{2}+{\left(26-14\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{0}^{2}+{12}^{2}}\phantom{\rule{0ex}{0ex}}=12units$

Suppose *d* be the total distance travelled by ayush

$d={d}_{1}+{d}_{2}+{d}_{3}\phantom{\rule{0ex}{0ex}}d=5+10+12\phantom{\rule{0ex}{0ex}}d=27km$

Now, let D be the shortest distance between ayush house and the office,

$D=\sqrt{{\left(13-2\right)}^{2}+{\left(26-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{11}^{2}+{22}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{605}\phantom{\rule{0ex}{0ex}}=24.6km$

Thus the extra distance covered by ayush is *d* − D = 27 − 24.6 = 2.4 km.

#### Page No 6.16:

#### Question 24:

Find the value of *k*, if the point *P* (0, 2) is equidistant from (3, *k*) and (*k*, 5).

#### Answer:

The distance *d* between two points and is given by the formula

It is said that *P*(0*,*2) is equidistant from both *A*(3*,k*) and *B*(*k,*5).

So, using the distance formula for both these pairs of points we have

Now since both these distances are given to be the same, let us equate both.

Squaring on both sides we have,

Hence the value of ‘*k*’ for which the point ‘*P*’ is equidistant from the other two given points is.

#### Page No 6.16:

#### Question 25:

If $\left(-4,3\right)\mathrm{and}(4,3)$ are two vertices of an equilateral triangle , find the coordinates of the third vertex , given that the origin lies in the

(i) interior (ii) exterior of the triangle .

#### Answer:

Suppose A(−4, 3) and B(4, 3) be the vertices of an equilateral triangle.

The distance AB is

$AB=\sqrt{\left(4-\left(-4\right)\right)+{\left(3-3\right)}^{2}}=8units$

Consider the figure below:

Here, D is point such that CD is perpendicular to AB.

Thus, AD = 4 units.

By pythagoras theorem, we have

$A{D}^{2}+C{D}^{2}=A{C}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {4}^{2}+C{D}^{2}={8}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow C{D}^{2}=64-16\phantom{\rule{0ex}{0ex}}\Rightarrow C{D}^{2}=48\phantom{\rule{0ex}{0ex}}\Rightarrow CD=4\sqrt{3}$

Since C is on the negative *y *axis, so coordinate of C are $3-4\sqrt{3}$ units.

If C is on the positive *y* axis, so coordinate of C are $3+4\sqrt{3}$ units.

#### Page No 6.16:

#### Question 26:

Show that the points (−3, 2), (−5,−5), (2, −3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.

#### Answer:

The distance *d* between two points and is given by the formula

In a rhombus all the sides are equal in length. And the area ‘*A*’ of a rhombus is given as

Here the four points are *A*(*−*3*,*2)*, B*(*−*5*,**−*5), *C*(2*,**−*3) and *D*(4*,*4)

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a.

Now let us find out the lengths of the diagonals of the rhombus.

Now using these values in the formula for the area of a rhombus we have,

Thus the area of the given rhombus is.

#### Page No 6.16:

#### Question 27:

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (−1, −6) and (4, −1). Also, find its circumradius.

#### Answer:

The distance *d* between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be *A*(3*,*0), *B*(*−*1*,**−*6) and *C*(4*,**−*1)

Let the circumcentre of the triangle be represented by the point *R*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co-ordinates of the circumcentre of the triangle with the given vertices are.

The length of the circumradius can be found out substituting the values of ‘*x*’ and ‘*y*’ in ‘*AR*’

Thus the circumradius of the given triangle is.

#### Page No 6.16:

#### Question 28:

Find a point on the *x*-axis which is equidistant from the points (7, 6) and (−3, 4).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x*−axis which is equidistant from both the points *A*(7*,*6) and *B*(*−*3*,*4).

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 6.16:

#### Question 29:

(i) Show that the points *A*(5, 6), *B*(1, 5), *C*(2, 1) and *D*(6,2) are the vertices of a square.

(ii) Prove hat the points *A* (2, 3) *B*(−2,2) *C*(−1,−2), and *D*(3, −1) are the vertices of a square *ABCD*.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are equal to each other. And also the diagonals are also equal to each other.

Here the four points are *A*(5*,*6)*, B*(1*,*5), *C*(2*,*1) and *D*(6*,*2).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Now let us find out the lengths of the diagonals of this rhombus.

Now since the diagonals of the rhombus are also equal to each other this rhombus has to be a square.

Hence we have proved that the quadrilateral formed by the given four points is a.

#### Page No 6.16:

#### Question 30:

Find the point on x-axis which is equidistant from the points (−2, 5) and (2,−3).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x-*axis which is equidistant from both the points *A*(*−*2*,*5) and *B*(2*,**−*3)

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 6.16:

#### Question 31:

Find the value of *x* such that *PQ* = *QR* where the coordinates of *P*,* Q* and *R* are (6, −1) , (1, 3) and (*x*, 8) respectively.

#### Answer:

The distance *d* between two points and is given by the formula

The three given points are *P*(6*,**−*1)*, Q*(1*,*3)* *and *R*(*x,*8).

Now let us find the distance between ‘*P*’ and ‘*Q*’.

Now, let us find the distance between ‘*Q*’ and ‘*R*’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Now we have a quadratic equation. Solving for the roots of the equation we have,

Thus the roots of the above equation are 5 and −3.

Hence the values of ‘*x*’ are.

#### Page No 6.16:

#### Question 32:

Prove that the points (0, 0), (5, 5) and (−5, 5) are the vertices of a right isosceles triangle.

#### Answer:

The distance *d* between two points and is given by the formula

In an isosceles triangle there are two sides which are equal in length.

By Pythagoras Theorem in a right-angled triangle the square of the longest side will be equal to the sum of squares of the other two sides.

Here the three points are *A*(0*,*0)*, B*(5*,*5) and *C*(*−*5*,*5).

Let us check the length of the three sides of the triangle.

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

Further it is seen that

If in a triangle the square of the longest side is equal to the sum of squares of the other two sides then the triangle has to be a right-angled triangle.

Hence proved that the triangle formed by the three given points is an.

#### Page No 6.16:

#### Question 33:

If the point *P*(*x*, *y*) is equidistant from the points *A*(5, 1) and *B* (1, 5), prove that *x* = *y*.

#### Answer:

The distance *d* between two points and is given by the formula

The three given points are* P*(*x, y*)*, A*(5*,*1)* *and *B*(1*,*5).

Now let us find the distance between ‘*P*’ and ‘*A*’.

Now, let us find the distance between ‘*P*’ and ‘*B*’.

$PB=\sqrt{{\left(x-1\right)}^{2}+{\left(y-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

It is given that both these distances are equal. So, let us equate both the above equations,

PA = PB

$\sqrt{{\left(x-5\right)}^{2}+{\left(y-1\right)}^{2}}=\sqrt{{\left(x-1\right)}^{2}+{\left(y-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

Squaring on both sides of the equation we get,

${\left(x-5\right)}^{2}+{\left(y-1\right)}^{2}={\left(x-1\right)}^{2}+{\left(y-5\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+25-10x+{y}^{2}+1-2y={x}^{2}+1-2x+{y}^{2}+25-10y\phantom{\rule{0ex}{0ex}}\Rightarrow 26-10x-2y=26-10y-2x\phantom{\rule{0ex}{0ex}}\Rightarrow 10y-2y=10x-2x\phantom{\rule{0ex}{0ex}}\Rightarrow 8y=8x\phantom{\rule{0ex}{0ex}}\Rightarrow y=x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence we have proved that *x** *= *y*.

#### Page No 6.16:

#### Question 34:

If *Q* (0,1) is equidistant from *P* (5, −3) and *R* (*x*, 6), find the values of *x*. Also, find the distances *QR* and *PR*.

#### Answer:

The distance *d* between two points and is given by the formula

The three given points are* Q*(0*,*1)*, P*(5*,**−*3)* *and *R*(*x,*6).

Now let us find the distance between ‘*P*’ and ‘*Q*’.

Now, let us find the distance between ‘*Q*’ and ‘*R*’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Hence the values of ‘*x*’ are.

Now, the required individual distances,

Hence the length of ‘*QR*’ is.

For ‘*PR*’ there are two cases. First when the value of ‘*x*’ is 4,

Then when the value of ‘*x*’ is −4,

Hence the length of ‘*PR*’ can beunits

#### Page No 6.16:

#### Question 35:

Find the values of *y* for which the distance between the points *P* (2, −3) and *Q*(10,*y*) is 10 units.

#### Answer:

The distance *d* between two points and is given by the formula

The distance between two points *P*(2*,**−*3) and *Q*(10*,y*) is given as 10 units. Substituting these values in the formula for distance between two points we have,

Now, squaring the above equation on both sides of the equals sign

Thus we arrive at a quadratic equation. Let us solve this now,

The roots of the above quadratic equation are thus 3 and −9.

Thus the value of ‘*y*’ could either be.

#### Page No 6.16:

#### Question 36:

If the point P(*k *− 1, 2) is equidistant from the points A(3, *k*) and B(*k*, 5), find the values of *k*. [CBSE 2014]

#### Answer:

It is given that P(*k *− 1, 2) is equidistant from the points A(3, *k*) and B(*k*, 5).

∴ AP = BP

$\Rightarrow \sqrt{{\left[\left(k-1\right)-3\right]}^{2}+{\left(2-k\right)}^{2}}=\sqrt{{\left[\left(k-1\right)-k\right]}^{2}+{\left(2-5\right)}^{2}}\left(\mathrm{Distance}\mathrm{formula}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{\left(k-4\right)}^{2}+{\left(2-k\right)}^{2}}=\sqrt{{\left(-1\right)}^{2}+{\left(-3\right)}^{2}}$

Squaring on both sides, we get

${k}^{2}-8k+16+4-4k+{k}^{2}=10\phantom{\rule{0ex}{0ex}}\Rightarrow 2{k}^{2}-12k+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}-6k+5=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(k-1\right)\left(k-5\right)=0$

$\Rightarrow k-1=0\mathrm{or}k-5=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=1\mathrm{or}k=5$

Thus, the value of *k* is 1 or 5.

#### Page No 6.17:

#### Question 37:

If the point A(0, 2) is equidistant from the points B(3, *p*) and C(*p*, 5), find *p*. Also, find the length of AB. [CBSE 2014]

#### Answer:

It is given that A(0, 2) is equidistant from the points B(3, *p*) and C(*p*, 5).

∴ AB = AC

$\Rightarrow \sqrt{{\left(3-0\right)}^{2}+{\left(p-2\right)}^{2}}=\sqrt{{\left(p-0\right)}^{2}+{\left(5-2\right)}^{2}}$ (Distance formula)

Squaring on both sides, we get

$9+{p}^{2}-4p+4={p}^{2}+9\phantom{\rule{0ex}{0ex}}\Rightarrow -4p+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=1$

Thus, the value of *p* is 1.

∴ AB = $\sqrt{{\left(3-0\right)}^{2}+{\left(1-2\right)}^{2}}=\sqrt{{3}^{2}+{\left(-1\right)}^{2}}=\sqrt{9+1}=\sqrt{10}$ units

#### Page No 6.17:

#### Question 38:

Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:

(i) A(−1,−2) B(1, 0), C (−1, 2), D(−3, 0)

(ii) A(−3, 5) B(3, 1), C (0, 3), D(−1, −4)

(iii) A(4, 5) B(7, 6), C (4, 3), D(1, 2)

#### Answer:

(i) A (−1,−2) , B(1,0), C(−1,2), D(−3,0)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

Hence

Similarly,

Similarly,

Also,

Hence from above we see that all the sides of the quadrilateral are equal. Hence it is a square.

(ii) A (−3,5) , B(3,1), C(0,3), D(−1,−4)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

Hence

Similarly,

Similarly,

Also,

Hence from the above we see that it is not a quadrilateral

(iii) A (4, 5), B (7,6), C(4,3), D(1,2)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points Pand Qis given by distance formula:

Hence

Similarly,

Similarly,

Also,

Hence from above we see that

AB = CD and BC = DA

Here opposite sides of the quadrilateral is equal. Hence it is a parallelogram.

#### Page No 6.17:

#### Question 39:

Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).

#### Answer:

TO FIND: The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(*x*, *y*) be any point on the perpendicular bisector of AB. Then,

PA=PB

Hence the equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5) is

#### Page No 6.17:

#### Question 40:

Prove that the points (3, 0), (4, 5), (−1, 4) and (−2 −1), taken in order, form a rhombus. Also, find its area.

#### Answer:

The distance *d* between two points and is given by the formula

In a rhombus all the sides are equal in length. And the area ‘*A*’ of a rhombus is given as

Here the four points are *A*(3*,*0)*, B*(4*,*5), *C*(*−*1*,*4) and *D*(*−*2*,**−*1).

First let us check if all the four sides are equal.

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a.

Now let us find out the lengths of the diagonals of the rhombus.

Now using these values in the formula for the area of a rhombus we have,

Thus the area of the given rhombus is.

#### Page No 6.17:

#### Question 41:

In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A(3, 1), B(6, 4), and C(8, 6). Do you think they are seated in a line?

#### Answer:

The distance *d* between two points and is given by the formula

For three points to be collinear the sum of distances between any two pairs of points should be equal to the third pair of points.

The given points are *A*(3*,*1), *B*(6*,*4) and *C*(8*,*6).

Let us find the distances between the possible pairs of points.

We see that.

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence, the three given points are.

#### Page No 6.17:

#### Question 42:

Find a point on *y*-axis which is equidistant form the points (5, −2) and (−3, 2).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the y−axis which is equidistant from both the points *A*(5*,**−*2) and *B*(*−*3*,*2).

Let this point be denoted as *C*(*x, y*).

Since the point lies on the *y*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘*A*’ and ‘*B*’ to ‘*C*’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *y*-axis which lies at equal distances from the mentioned points is.

#### Page No 6.17:

#### Question 43:

Find a relation between *x* and *y* such that the point (*x*, *y*) is equidistant from the points (3, 6) and (−3, 4).

#### Answer:

The distance *d* between two points and is given by the formula

Let the three given points be* P*(*x, y*)*, A*(3*,*6)* *and *B*(*−*3*,*4).

Now let us find the distance between ‘*P*’ and ‘*A*’.

Now, let us find the distance between ‘*P*’ and ‘*B*’.

It is given that both these distances are equal. So, let us equate both the above equations,

Squaring on both sides of the equation we get,

Hence the relationship between ‘*x*’ and ‘*y*’ based on the given condition is.

#### Page No 6.17:

#### Question 44:

If a point A(0, 2) is equidistant from the points B(3, *p*) and C(*p*, 5), then find the value of *p*. [CBSE 2012, 2013]

#### Answer:

It is given that A(0, 2) is equidistant from the points B(3, *p*) and C(*p*, 5).

∴ AB = AC

$\Rightarrow \sqrt{{\left(3-0\right)}^{2}+{\left(p-2\right)}^{2}}=\sqrt{{\left(p-0\right)}^{2}+{\left(5-2\right)}^{2}}$ (Distance formula)

Squaring on both sides, we get

$9+{p}^{2}-4p+4={p}^{2}+9\phantom{\rule{0ex}{0ex}}\Rightarrow -4p+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=1$

Thus, the value of *p* is 1.

#### Page No 6.17:

#### Question 45:

Prove that the points (7, 10), (−2, 5) and (3, −4) are the vertices of an isosceles right triangle. [CBSE 2013]

#### Answer:

Let the given points be A(7, 10), B(−2, 5) and C(3, −4).

Using distance formula, we have

$\mathrm{AB}=\sqrt{{\left(-2-7\right)}^{2}+{\left(5-10\right)}^{2}}=\sqrt{{\left(-9\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{81+25}=\sqrt{106}$ units

$\mathrm{BC}=\sqrt{{\left[3-\left(-2\right)\right]}^{2}+{\left(-4-5\right)}^{2}}=\sqrt{{5}^{2}+{\left(-9\right)}^{2}}=\sqrt{25+81}=\sqrt{106}$ units

$\mathrm{CA}=\sqrt{{\left(3-7\right)}^{2}+{\left(-4-10\right)}^{2}}=\sqrt{{\left(-4\right)}^{2}+{\left(-14\right)}^{2}}=\sqrt{16+196}=\sqrt{212}$ units

Thus, AB = BC = $\sqrt{106}$ units

∴ ∆ABC is an isosceles triangle.

Also,

AB^{2} + BC^{2 }= 106 + 106 = 212

and CA^{2} = 212

∴ AB^{2} + BC^{2 }= CA^{2}

So, ∆ABC is right angled at B. (Converse of Pythagoras theorem)

Hence, the given points are the vertices of an isosceles right triangle.

#### Page No 6.17:

#### Question 46:

If the point P(*x*, 3) is equidistant from the point A(7, −1) and B(6, 8), then find the value of *x* and find the distance AP. [CBSE 2014]

#### Answer:

It is given that P(*x*, 3) is equidistant from the point A(7, −1) and B(6, 8).

∴ AP = BP

$\Rightarrow \sqrt{{\left(x-7\right)}^{2}+{\left[3-\left(-1\right)\right]}^{2}}=\sqrt{{\left(x-6\right)}^{2}+{\left(8-3\right)}^{2}}$ (Distance formula)

Squaring on both sides, we get

${\left(x-7\right)}^{2}+16={\left(x-6\right)}^{2}+25\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-14x+49+16={x}^{2}-12x+36+25\phantom{\rule{0ex}{0ex}}\Rightarrow -14x+12x=61-65\phantom{\rule{0ex}{0ex}}\Rightarrow -2x=-4\phantom{\rule{0ex}{0ex}}\Rightarrow x=2$

Thus, the value of *x* is 2.

$\therefore \mathrm{AP}=\sqrt{{\left(2-7\right)}^{2}+{\left[3-\left(-1\right)\right]}^{2}}=\sqrt{{\left(-5\right)}^{2}+{4}^{2}}=\sqrt{25+16}=\sqrt{41}$ units

#### Page No 6.17:

#### Question 47:

If A(3, *y*) is equidistant from points P(8, −3) and Q(7, 6), find the value of *y* and find the distance AQ. [CBSE 2014]

#### Answer:

It is given that A(3, *y*) is equidistant from points P(8, −3) and Q(7, 6).

∴ AP = AQ

$\Rightarrow \sqrt{{\left(3-8\right)}^{2}+{\left[y-\left(-3\right)\right]}^{2}}=\sqrt{{\left(3-7\right)}^{2}+{\left(y-6\right)}^{2}}$ (Distance formula)

Squaring on both sides, we get

$25+{\left(y+3\right)}^{2}=16+{\left(y-6\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25+{y}^{2}+6y+9=16+{y}^{2}-12y+36\phantom{\rule{0ex}{0ex}}\Rightarrow 12y+6y=52-34\phantom{\rule{0ex}{0ex}}\Rightarrow 18y=18\phantom{\rule{0ex}{0ex}}\Rightarrow y=1$

Thus, the value of *y* is 1.

$\therefore \mathrm{AQ}=\sqrt{{\left(3-7\right)}^{2}+{\left(1-6\right)}^{2}}=\sqrt{{\left(-4\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{16+25}=\sqrt{41}$ units

#### Page No 6.17:

#### Question 48:

If (0, −3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. [CBSE 2014]

#### Answer:

Let the given points be A(0, −3) and B(0, 3). Suppose the coordinates of the third vertex be C(*x*, *y*).

Now, ∆ABC is an equilateral triangle.

∴ AB = BC = CA

$\sqrt{{\left(0-0\right)}^{2}+{\left(-3-3\right)}^{2}}=\sqrt{{\left(x-0\right)}^{2}+{\left(y-3\right)}^{2}}=\sqrt{{\left(x-0\right)}^{2}+{\left[y-\left(-3\right)\right]}^{2}}$ (Distance formula)

Squaring on both sides, we get

$36={x}^{2}+{\left(y-3\right)}^{2}={x}^{2}+{\left(y+3\right)}^{2}$

⇒ ${x}^{2}+{\left(y-3\right)}^{2}={x}^{2}+{\left(y+3\right)}^{2}$ and ${x}^{2}+{\left(y-3\right)}^{2}=36$

Now,

${x}^{2}+{\left(y-3\right)}^{2}={x}^{2}+{\left(y+3\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}^{2}-6y+9={y}^{2}+6y+9\phantom{\rule{0ex}{0ex}}\Rightarrow -12y=0\phantom{\rule{0ex}{0ex}}\Rightarrow y=0$

Putting *y* = 0 in ${x}^{2}+{\left(y-3\right)}^{2}=36$, we get

${x}^{2}+{\left(0-3\right)}^{2}=36\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=36-9=27\phantom{\rule{0ex}{0ex}}\Rightarrow x=\pm \sqrt{27}=\pm 3\sqrt{3}$

Thus, the coordinates of the third vertex are $\left(3\sqrt{3},0\right)$ or $\left(-3\sqrt{3},0\right)$.

#### Page No 6.17:

#### Question 49:

If the point P(2, 2) is equidistant from the points A(−2, *k*) and B(−2*k*, −3), find *k*. Also, find the length of AP. [CBSE 2014]

#### Answer:

It is given that P(2, 2) is equidistant from the points A(−2, *k*) and B(−2*k*, −3).

∴ AP = BP

$\Rightarrow \sqrt{{\left(-2-2\right)}^{2}+{\left(k-2\right)}^{2}}=\sqrt{{\left(-2k-2\right)}^{2}+{\left(-3-2\right)}^{2}}$ (Distance formula)

Squaring on both sides, we get

$16+{k}^{2}-4k+4=4{k}^{2}+8k+4+25\phantom{\rule{0ex}{0ex}}\Rightarrow 3{k}^{2}+12k+9=0\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}+4k+3=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(k+3\right)\left(k+1\right)=0$

$\Rightarrow k+3=0\mathrm{or}k+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=-3\mathrm{or}k=-1$

Thus, the value of *k* is −1 or −3.

When *k* = −1,

$\mathrm{AP}=\sqrt{{\left(-2-2\right)}^{2}+{\left(-1-2\right)}^{2}}=\sqrt{{\left(-4\right)}^{2}+{\left(-3\right)}^{2}}=\sqrt{16+9}=\sqrt{25}=5$ units

When *k* = −3,

$\mathrm{AP}=\sqrt{{\left(-2-2\right)}^{2}+{\left(-3-2\right)}^{2}}=\sqrt{{\left(-4\right)}^{2}+{\left(-5\right)}^{2}}=\sqrt{16+25}=\sqrt{41}$ units

#### Page No 6.17:

#### Question 50:

Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 2) are similar triangles.

#### Answer:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2).

$AB=\sqrt{{\left(2+2\right)}^{2}+{\left(0-0\right)}^{2}}=4\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(0-2\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{8}=2\sqrt{2}\phantom{\rule{0ex}{0ex}}CA=\sqrt{{\left(0+2\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{8}=2\sqrt{2}$

In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).

$PQ=\sqrt{{\left(4+4\right)}^{2}+{\left(0-0\right)}^{2}}=8\phantom{\rule{0ex}{0ex}}QR=\sqrt{{\left(0-4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}\phantom{\rule{0ex}{0ex}}PR=\sqrt{{\left(0+4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}$

Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional.

$\mathrm{So},\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{PR}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4}{8}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$

Thus, ΔABC is similar to ΔPQR.

#### Page No 6.17:

#### Question 51:

An equilateral triangle has two vertices at the points (3, 4) and (−2, 3), find the coordinates of the third vertex.

#### Answer:

The distance *d* between two points and is given by the formula

In an equilateral triangle all the sides are of equal length.

Here we are given that *A *(3*, *4) and *B *(*−*2*, *3) are two vertices of an equilateral triangle. Let *C*(*x, y*) be the third vertex of the equilateral triangle.

First let us find out the length of the side of the equilateral triangle.

Hence the side of the equilateral triangle measures units.

Now, since it is an equilateral triangle, all the sides need to measure the same length.

Hence we have

Equating both these equations we have,

Squaring on both sides we have,

From the above equation we have,

Substituting this and the value of the side of the triangle in the equation for one of the sides we have,

Squaring on both sides,

Now we have a quadratic equation for ‘*x*’. Solving for the roots of this equation,

We know that. Substituting the value of ‘*x*’ we have,

Hence the two possible values of the third vertex are.

#### Page No 6.17:

#### Question 52:

Find the circumcentre of the triangle whose vertices are (−2, −3), (−1, 0), (7, −6).

#### Answer:

The distance *d* between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be *A*(*−*2*.−*3), *B*(*−*1*,*0) and *C*(7*,**−*6).

Let the circumcentre of the triangle be represented by the point *R*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co−ordinates of the circumcentre of the triangle with the given vertices are.

#### Page No 6.17:

#### Question 53:

Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0).

#### Answer:

The distance *d* between two points and is given by the formula

In a right angled triangle the angle opposite the hypotenuse subtends an angle of.

Here let the given points be *A*(0*,*100)*, B*(10*,*0). Let the origin be denoted by *O*(0*,*0).

Let us find the distance between all the pairs of points

Here we can see that.

So, is a right angled triangle with ‘*AB*’ being the hypotenuse. So the angle opposite it has to be. This angle is nothing but the angle subtended by the line segment ‘*AB*’ at the origin.

Hence the angle subtended at the origin by the given line segment is.

#### Page No 6.17:

#### Question 54:

Find the centre of the circle passing through (5, −8), (2, −9) and (2, 1).

#### Answer:

The distance *d* between two points and is given by the formula

The centre of a circle is at equal distance from all the points on its circumference.

Here it is given that the circle passes through the points *A*(5*,**−*8), *B*(2*,**−*9) and *C*(2*,*1).

Let the centre of the circle be represented by the point *O*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co-ordinates of the centre of the circle are.

#### Page No 6.17:

#### Question 55:

If two opposite vertices of a square are (5, 4) and (1, −6), find the coordinates of its remaining two vertices.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to times the side of the square.

Here let the two points which are said to be the opposite vertices of a diagonal of a square be *A*(5*,*4)* *and *C*(1*,**−*6).

Let us find the distance between them which is the length of the diagonal of the square.

Now we know that in a square,

Substituting the value of the diagonal we found out earlier in this equation we have,

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let *P*(*x, y*) represent another vertex of the same square adjacent to both ‘*A*’ and ‘*C*’.

But these two are nothing but the sides of the square and need to be equal to each other.

Squaring on both sides we have,

From this we have,

Substituting this value of ‘*x*’ and the length of the side in the equation for ‘*AP*’ we have,

Squaring on both sides,

We have a quadratic equation. Solving for the roots of the equation we have,

The roots of this equation are −3 and 1.

Now we can find the respective values of ‘*x*’ by substituting the two values of ‘*y*’

When

When

Therefore the other two vertices of the square are.

#### Page No 6.17:

#### Question 56:

Find the centre of the circle passing through (6, −6), (3, −7) and (3, 3).

#### Answer:

The distance *d* between two points and is given by the formula

The centre of a circle is at equal distance from all the points on its circumference.

Here it is given that the circle passes through the points *A*(6*,**−*6), *B*(3*,**−*7) and *C*(3*,*3).

Let the centre of the circle be represented by the point *O*(*x, y*).

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Now we have two equations for ‘*x*’ and ‘*y*’, which are

From the second equation we have. Substituting this value of ‘*y*’ in the first equation we have,

Therefore the value of ‘*y*’ is,

Hence the co-ordinates of the centre of the circle are.

#### Page No 6.17:

#### Question 57:

Two opposite vertices of a square are (−1, 2) and (3, 2). Find the coordinates of other two vertices.

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to times the side of the square.

Here let the two points which are said to be the opposite vertices of a diagonal of a square be *A*(*−*1*,*2)* *and *C*(3*,*2).

Let us find the distance between them which is the length of the diagonal of the square.

Now we know that in a square,

Substituting the value of the diagonal we found out earlier in this equation we have,

Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.

Let *P*(*x, y*) represent another vertex of the same square adjacent to both ‘*A*’ and ‘*C*’.

But these two are nothing but the sides of the square and need to be equal to each other.

Squaring on both sides we have,

Substituting this value of ‘*x*’ and the length of the side in the equation for ‘*AP*’ we have,

Squaring on both sides,

We have a quadratic equation. Solving for the roots of the equation we have,

The roots of this equation are 0 and 4.

Therefore the other two vertices of the square are.

#### Page No 6.28:

#### Question 1:

Find the coordinates of the point which divides the line segment joining (−1,3) and (4, −7) internally in the ratio 3 : 4

#### Answer:

We have A (−1, 3) and B (4,−7) be two points. Let a pointdivide the line segment joining the points A and B in the ratio 3:4 internally.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula to find the co-ordinates of unknown point P as,

Therefore, co-ordinates of point P is

#### Page No 6.28:

#### Question 2:

Find the points of trisection of the line segment joining the points:

(a) 5, −6 and (−7, 5),

(b) (3, −2) and (−3, −4),

(c) (2, −2) and (−7, 4).

#### Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

The points of trisection of a line are the points which divide the line into the ratio.

(i) Here we are asked to find the points of trisection of the line segment joining the points *A*(5*,*−6) and *B*(−7*,*5).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1 : 2.

Let *Q*(*e, d*) be the point which divides the line joining ‘*AB*’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are .

(ii) Here we are asked to find the points of trisection of the line segment joining the points *A*(3*,*−2) and *B*(−3*,*−4).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1 : 2.

Let *Q*(*e, d*) be the point which divides the line joining ‘*AB*’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are.

(iii) Here we are asked to find the points of trisection of the line segment joining the points *A*(2*,*−2) and *B*(−7*,*4).

So we need to find the points which divide the line joining these two points in the ratio and 2 : 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1 : 2.

Let *Q*(*e, d*) be the point which divides the line joining ‘*AB*’ in the ratio 2 : 1.

Therefore the points of trisection of the line joining the given points are .

#### Page No 6.28:

#### Question 3:

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (−2, −1), (1, 0), (4, 3) and(1, 2) meet.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

In a parallelogram the diagonals bisect each other. That is the point of intersection of the diagonals is the midpoint of either of the diagonals.

Here, it is given that the vertices of a parallelogram are *A*(−2*,*−1)*, B*(1*,*0) and *C*(4*,*3) and *D*(1*,*2).

We see that ‘*AC*’ and ‘*BD*’ are the diagonals of the parallelogram.

The midpoint of either one of these diagonals will give us the point of intersection of the diagonals.

Let this point be *M*(*x, y*).

Let us find the midpoint of the diagonal ‘*AC*’.

Hence the co-ordinates of the point of intersection of the diagonals of the given parallelogram are.

#### Page No 6.28:

#### Question 4:

Prove that the points (3, −2), (4, 0), (6, −3) and (5, −5) are the vertices of a parallelogram.

#### Answer:

Let A (3,−2); B (4, 0); C (6,−3) and D (5,−5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

#### Page No 6.28:

#### Question 5:

If *P *( 9*a* $-$2 ,$-$b) divides the line segment joining *A *(3*a* + 1 , $-$3 ) and* B *(8*a*, 5) in the ratio 3 : 1 , find the values of *a* and *b *.

#### Answer:

It is given that P divides AB in the ratio 3 : 1.

Therefore, by section formula we have

$\Rightarrow 9a-2=\frac{3\left(8a\right)+1\left(3a+1\right)}{3+1}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\left(9a-2\right)=24a+3a+1\phantom{\rule{0ex}{0ex}}\Rightarrow 36a-8=27a+1\phantom{\rule{0ex}{0ex}}\Rightarrow 9a=9\phantom{\rule{0ex}{0ex}}\Rightarrow a=1$

And,

$\Rightarrow -b=\frac{3\left(5\right)+1\left(-3\right)}{3+1}\phantom{\rule{0ex}{0ex}}\Rightarrow -4b=15-3\phantom{\rule{0ex}{0ex}}\Rightarrow b=-3$

#### Page No 6.28:

#### Question 6:

If (*a*,*b*) is the mid-point of the line segment joining the points *A *(10, $-$6) , *B *(*k*,4) and *a*$-$2*b *= 18 , find the value of *k* and the distance *AB.*

#### Answer:

It is given that A(10, −6) and B(*k*, 4).

Suppose (*a*, *b*) be midpoint of AB. Then,

$a=\frac{10+k}{2},b=\frac{-6+4}{2}=\frac{-2}{2}=-1\phantom{\rule{0ex}{0ex}}Now,a-2b=18\phantom{\rule{0ex}{0ex}}\Rightarrow a=18-2=16\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}16\times 2=10+k\phantom{\rule{0ex}{0ex}}\Rightarrow k=22\phantom{\rule{0ex}{0ex}}Further,\phantom{\rule{0ex}{0ex}}AB=\sqrt{{\left(22-10\right)}^{2}+{\left(4+6\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{144+100}\phantom{\rule{0ex}{0ex}}=2\sqrt{61}$

#### Page No 6.29:

#### Question 7:

Find the ratio in which the point (2, *y*) divides the line segment joining the points *A* (−2, 2) and *B* (3, 7). Also, find the value of *y*.

#### Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here we are given that the point *P*(2*,y*) divides the line joining the points *A*(−2*,*2) and *B*(3*,*7) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘*y*’.

Equating the individual components we have

Thus the value of ‘*y*’ is.

#### Page No 6.29:

#### Question 8:

If *A* (−1, 3), *B* (1, −1) and *C* (5, 1) are the vertices of a triangle *ABC*, find the length of the median through* A*.

#### Answer:

The distance *d* between two points and is given by the formula

The co-ordinates of the midpoint between two points and is given by,

Here, it is given that the three vertices of a triangle are *A*(−1*,*3)*, B*(1*,*−1) and *C*(5*,*1).

The median of a triangle is the line joining a vertex of a triangle to the mid-point of the side opposite this vertex.

Let ‘*D*’ be the mid-point of the side ‘*BC*’.

Let us now find its co-ordinates.

Thus we have the co-ordinates of the point as *D*(3*,*0).

Now, let us find the length of the median ‘*AD*’.

Thus the length of the median through the vertex ‘*A*’ of the given triangle is.

#### Page No 6.29:

#### Question 9:

If the points P, Q(*x*, 7), R, S(6, *y*) in this order divide the line segment joining A(2, *p*) and B(7, 10) in 5 equal parts, find *x*, *y* and *p*. [CBSE 2015]* *

#### Answer:

It is given that P, Q(*x*, 7), R, S(6, *y*) divides the line segment joining A(2, *p*) and B(7, 10) in 5 equal parts.

∴ AP = PQ = QR = RS = SB .....(1)

Now,

AP + PQ + QR + RS + SB = AB

⇒ SB + SB + SB + SB + SB = AB [From (1)]

⇒ 5SB = AB

⇒ SB = $\frac{1}{5}$AB .....(2)

Now,

AS = AP + PQ + QR + RS = $\frac{1}{5}$AB +$\frac{1}{5}$AB + $\frac{1}{5}$AB + $\frac{1}{5}$AB = $\frac{4}{5}$AB .....(3)

From (2) and (3), we get

AS : SB = $\frac{4}{5}$AB : $\frac{1}{5}$AB = 4 : 1

Similarly,

AQ : QB = 2 : 3

Using section formula, we get

Coordinates of Q = $\left(\frac{2\times 7+3\times 2}{2+3},\frac{2\times 10+3\times p}{2+3}\right)=\left(\frac{20}{5},\frac{20+3p}{5}\right)=\left(4,\frac{20+3p}{5}\right)$

$\therefore \left(x,7\right)=\left(4,\frac{20+3p}{5}\right)$

$\Rightarrow x=4$ and $7=\frac{20+3p}{5}$

Now,

$7=\frac{20+3p}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 20+3p=35\phantom{\rule{0ex}{0ex}}\Rightarrow 3p=15\phantom{\rule{0ex}{0ex}}\Rightarrow p=5$

Coordinates of S = $\left(\frac{4\times 7+1\times 2}{4+1},\frac{4\times 10+1\times p}{4+1}\right)=\left(\frac{30}{5},\frac{40+5}{5}\right)=\left(6,9\right)$

$\therefore \left(6,y\right)=\left(6,9\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y=9$

Thus, the values of *x*, *y* and *p* are 4, 9 and 5, respectively.

#### Page No 6.29:

#### Question 10:

If a vertex of a triangle be (1, 1) and the middle points of the sides through it be (−2,−3) and (5 2) find the other vertices.

#### Answer:

Let a in which P and Q are the mid-points of sides AB and AC respectively. The coordinates are: A (1, 1); P (−2, 3) and Q (5, 2).

We have to find the co-ordinates of and.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So, co-ordinates of B is (−5, 5)

Similarly, mid-point Q of side AC can be written as,

Now equate the individual terms to get,

So, co-ordinates of C is (9, 3)

#### Page No 6.29:

#### Question 11:

(i) In what ratio is the line segment joining the points (−2,−3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

(ii) In what ratio is the line segment joining (−3, −1) and (−8, −9) divided at the point (−5, −21/5)?

#### Answer:

(i) The ratio in which the y-axis divides two points and is $\lambda :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

; where

Here the two given points are *A*(−2*,*−3) and *B*(3*,*7).

Since, the point is on the y-axis so, *x* coordinate is 0.

$\frac{3\lambda -2}{1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{2}{3}$

Thus the given points are divided by the y-axis in the ratio.

The co-ordinates of this point (*x, y*)* *can be found by using the earlier mentioned formula.

Thus the co-ordinates of the point which divides the given points in the required ratio are.

(ii) The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here it is said that the point divides the points (−3*,*−1) and (−8*,*−9). Substituting these values in the above formula we have,

Equating the individual components we have,

Therefore the ratio in which the line is divided is.

#### Page No 6.29:

#### Question 12:

If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of *k*.

#### Answer:

We have two points A (3, 4) and B (k, 7) such that its mid-point is.

It is also given that point P lies on a line whose equation is

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

Since, P lies on the given line. So,

Put the values of co-ordinates of point P in the equation of line to get,

On further simplification we get,

So,

#### Page No 6.29:

#### Question 13:

Find the ratio in which the point P$\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points A$\left(\frac{1}{2},\frac{3}{2}\right)$ and B$\left(2,-5\right)$. [CBSE 2015]

#### Answer:

Suppose P$\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points A$\left(\frac{1}{2},\frac{3}{2}\right)$ and B$\left(2,-5\right)$ in the ratio *k* : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{2k+{\displaystyle \frac{1}{2}}}{k+1},\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}\right)$

$\therefore \left(\frac{2k+{\displaystyle \frac{1}{2}}}{k+1},\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}\right)=\left(\frac{3}{4},\frac{5}{12}\right)$

$\Rightarrow \frac{2k+{\displaystyle \frac{1}{2}}}{k+1}=\frac{3}{4}$ and $\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}=\frac{5}{12}$

Now,

$\frac{2k+{\displaystyle \frac{1}{2}}}{k+1}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 8k+2=3k+3\phantom{\rule{0ex}{0ex}}\Rightarrow 5k=1\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{1}{5}$

Putting *k* = $\frac{1}{5}$ in $\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}=\frac{5}{12}$, we get

LHS = $\frac{-5\times {\displaystyle \frac{1}{5}}+{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{1}{5}}+1}=\frac{-1+{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{1}{5}}+1}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{6}{5}}}=\frac{5}{12}$ = RHS

Thus, the required ratio is $\frac{1}{5}$ : 1 or 1 : 5.

#### Page No 6.29:

#### Question 14:

Find the ratio in which the line segment joining (−2, −3) and (5, 6) is divided by (i) *x*-axis (ii) *y*-axis. Also, find the coordinates of the point of division in each case.

#### Answer:

The ratio in which the *x*−axis divides two points and is $\lambda :1$

The ratio in which the y-axis divides two points and is $\mu :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

Where

Here the two given points are *A*(−2*,*−3) and *B*(5*,*6).

- The ratio in which the x-axis divides these points is

$\frac{6\lambda -3}{3}=0\phantom{\rule{0ex}{0ex}}\lambda =\frac{1}{2}$

Let point *P*(*x, y*) divide the line joining ‘*AB*’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the ratio in which the *x*−axis divides the two given points and the co-ordinates of the point is.

- The ratio in which the y-axis divides these points is

$\frac{5\mu -2}{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =\frac{2}{5}$

Let point *P*(*x, y*) divide the line joining ‘*AB*’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the ratio in which the x-axis divides the two given points and the co-ordinates of the point is.

#### Page No 6.29:

#### Question 15:

Prove that the points (4, 5) (7, 6), (6, 3) (3, 2) are the vertices of a parallelogram. Is it a rectangle.

#### Answer:

Let A (4, 5); B (7, 6); C (6, 3) and D (3, 2) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

Now to check if ABCD is a rectangle, we should check the diagonal length.

Similarly,

Diagonals are of different lengths.

Hence ABCD is not a rectangle.

#### Page No 6.29:

#### Question 16:

Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square.

#### Answer:

Let A (4, 3); B (6, 4); C (5, 6) and D (3, 5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a square.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

All the sides as well as the diagonals are equal. Hence ABCD is a square.

#### Page No 6.29:

#### Question 17:

Prove that the points (−4,−1), (−2, 4), (4, 0) and (2, 3) are the vertices of a rectangle.

#### Answer:

Let A (−4,−1); B (−2,−4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.

So we should find the lengths of opposite sides of quadrilateral ABCD.

Opposite sides are equal. So now we will check the lengths of the diagonals.

Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.

#### Page No 6.29:

#### Question 18:

Find the lengths of the medians of a triangle whose vertices are A (−1,3), B(1,−1) and C(5, 1).

#### Answer:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (−1, 3); B (1,−1) and C (5, 1).

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (0, 1)

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (3, 0)

Similarly mid-point R of side AC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (2, 2)

Therefore length of median from A to the side BC is,

Similarly length of median from B to the side AC is,

Similarly length of median from C to the side AB is

#### Page No 6.29:

#### Question 19:

Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by the *x*-axis. Also, find the coordinates of the point of division. [CBSE 2014]

#### Answer:

Suppose the *x*-axis divides the line segment joining the points A(3, −3) and B(−2, 7) in the ratio *k* : 1.

Using section formula, we get

Coordinates of the point of division = $\left(\frac{-2k+3}{k+1},\frac{7k-3}{k+1}\right)$

Since the point of division lies on the *x*-axis, so its *y*-coordinate is 0.

$\therefore \frac{7k-3}{k+1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 7k-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{3}{7}$

So, the required ratio is $\frac{3}{7}$ : 1 or 3 : 7.

Putting *k* = $\frac{3}{7}$, we get

Coordinates of the point of division = $\left(\frac{-2\times {\displaystyle \frac{3}{7}}+3}{{\displaystyle \frac{3}{7}}+1},0\right)=\left(\frac{-6+21}{3+7},0\right)=\left(\frac{15}{10},0\right)=\left(\frac{3}{2},0\right)$

Thus, the coordinates of the point of division are $\left(\frac{3}{2},0\right)$.

#### Page No 6.29:

#### Question 20:

Find the ratio in which the point P(*x*, 2) divides the line segment joining the points A(12, 5) and B(4, −3). Also, find the value of *x*. [CBSE 2014]

#### Answer:

Suppose P(*x*, 2) divides the line segment joining the points A(12, 5) and B(4, −3) in the ratio *k* : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{4k+12}{k+1},\frac{-3k+5}{k+1}\right)$

$\therefore \left(\frac{4k+12}{k+1},\frac{-3k+5}{k+1}\right)=\left(x,2\right)$

$\Rightarrow x=\frac{4k+12}{k+1}$ and $\frac{-3k+5}{k+1}=2$

Now,

$\frac{-3k+5}{k+1}=2\phantom{\rule{0ex}{0ex}}\Rightarrow -3k+5=2k+2\phantom{\rule{0ex}{0ex}}\Rightarrow 5k=3\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{3}{5}$

So, P divides the line segment AB in the ratio 3 : 5.

Putting *k* = $\frac{3}{5}$ in $x=\frac{4k+12}{k+1}$, we get

$x=\frac{4\times {\displaystyle \frac{3}{5}}+12}{{\displaystyle \frac{3}{5}}+1}=\frac{12+60}{3+5}=\frac{72}{8}=9$

Thus, the value of *x* is 9.

#### Page No 6.29:

#### Question 21:

Find the ratio in which the point P(−1, *y*) lying on the line segment joining A(−3, 10) and B(6 −8) divides it. Also find the value of *y*. [CBSE 2013]

#### Answer:

Suppose P(−1, *y*) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio *k* : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$

$\therefore \left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)=\left(-1,y\right)$

$\Rightarrow \frac{6k-3}{k+1}=-1$ and $y=\frac{-8k+10}{k+1}$

Now,

$\frac{6k-3}{k+1}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow 6k-3=-k-1\phantom{\rule{0ex}{0ex}}\Rightarrow 7k=2\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{2}{7}$

So, P divides the line segment AB in the ratio 2 : 7.

Putting *k* = $\frac{2}{7}$ in $y=\frac{-8k+10}{k+1}$, we get

$y=\frac{-8\times {\displaystyle \frac{2}{7}}+10}{{\displaystyle \frac{2}{7}}+1}=\frac{-16+70}{2+7}=\frac{54}{9}=6$

Hence, the value of *y* is 6.

#### Page No 6.29:

#### Question 22:

Find the coordinates of a point *A*, where *AB* is a diameter of the circle whose centre is (2, −3) and *B* is (1, 4).

#### Answer:

Let the co-ordinates of point A be.

Centre lies on the mid-point of the diameter. So applying the mid-point formula we get,

Similarly,

So the co-ordinates of A are (3,−10)

#### Page No 6.29:

#### Question 23:

If the points (−2, −1), (1, 0), (*x*, 3) and (1, *y*) form a parallelogram, find the values of *x* and *y*.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−2,−1); B (1, 0); C (*x*, 3) and D (1, *y*).

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

#### Page No 6.29:

#### Question 24:

The points A(2, 0), B(9, 1) C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

#### Answer:

Let A (2, 0); B (9, 1); C (11, 6) and D (4, 4) be the vertices of a quadrilateral. We have to check if the quadrilateral ABCD is a rhombus or not.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are unequal. Hence ABCD is not a rhombus.

#### Page No 6.29:

#### Question 25:

In what ratio does the point (−4, 6) divide the line segment joining the points A(−6, 10) and B(3,−8)?

#### Answer:

The co-ordinates of a point which divided two points and internally in the ratio is given by the formula,

Here it is said that the point (−4*,*6)* *divides the points *A*(−6*,*10) and *B*(3*,*−8). Substituting these values in the above formula we have,

Equating the individual components we have,

Therefore the ratio in which the line is divided is

#### Page No 6.29:

#### Question 26:

Find the ratio in which the *y*-axis divides the line segment joining the points (5, −6) and (−1,−4). Also, find the coordinates of the point of division.

#### Answer:

The ratio in which the y-axis divides two points and is $\lambda :1$

The co-ordinates of the point dividing two points and in the ratio is given as,

where,

Here the two given points are *A*(5*,*−6) and *B*(−1*,*−4).

$(x,y)=\left(\frac{-\lambda +5}{\lambda +1},\frac{-4\lambda -6}{\lambda +1}\right)$

Since, the y-axis divided the given line, so the x coordinate will be 0.

$\frac{-\lambda +5}{\lambda +1}=0\phantom{\rule{0ex}{0ex}}\lambda =\frac{5}{1}$

Thus the given points are divided by the y-axis in the ratio.

The co-ordinates of this point (*x, y*)* *can be found by using the earlier mentioned formula.

Thus the co-ordinates of the point which divides the given points in the required ratio are.

#### Page No 6.29:

#### Question 27:

Show that *A* (−3, 2), *B* (−5, −5), *C *(2,−3), and *D* (4, 4) are the vertices of a rhombus.

#### Answer:

Let A (−3, 2); B (−5,−5); C (2,−3) and D (4, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rhombus.

So we should find the lengths of sides of quadrilateral ABCD.

All the sides of quadrilateral are equal. Hence ABCD is a rhombus.

#### Page No 6.29:

#### Question 28:

Find the length of the medians of a Δ*ABC* having vertices at A(0, −1), B(2, 1) and C(0, 3).

#### Answer:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,−1); B (2, 1) and C (0, 3).

So we should find the mid-points of the sides of the triangle.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (1, 0)

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (1, 2)

Similarly mid-point R of side AC can be written as,

Now equate the individual terms to get,

So co-ordinates of R is (0, 1)

Therefore length of median from A to the side BC is,

Similarly length of median from B to the side AC is,

Similarly length of median from C to the side AB is

#### Page No 6.29:

#### Question 29:

Find the ratio in which *P*(4, *m*) divides the line segment joining the points *A*(2, 3) and *B*(6, –3). Hence, find *m*.

#### Answer:

Let P divides AB in a ratio of *λ* : 1

Therefore, coordinates of the point P are $\left(\frac{6\lambda +2}{\lambda +1},\frac{-3\lambda +3}{\lambda +1}\right)$

Given that coordinates of the point P are (4, *m*).

$\Rightarrow \frac{6\lambda +2}{\lambda +1}=4\phantom{\rule{0ex}{0ex}}\Rightarrow 6\lambda +2=4\lambda +4\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =1$

Hence, the point P divides AB in a ratio of 1 : 1.

Replacing the value of *λ* = 1 in *y-*coordinate of P, we get

$\frac{-3\left(1\right)+3}{1+1}=m\phantom{\rule{0ex}{0ex}}\Rightarrow m=0$

Thus, *y*-coordinate of P is equal to 0.

#### Page No 6.30:

#### Question 30:

Find the coordinates of the points which divide the line segment joining the points (−4, 0) and (0, 6) in four equal parts.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Here we are supposed to find the points which divide the line joining *A*(−4*,*0) and *B*(0*,*6) into 4 equal parts.

We shall first find the midpoint *M*(*x, y*)* *of these two points since this point will divide the line into two equal parts

So the point *M*(−2*,*3) splits this line into two equal parts.

Now, we need to find the midpoint of *A*(−4*,*0) and *M*(−2*,*3) separately and the midpoint of *B*(0*,*6) and *M*(−2*,*3). These two points along with *M*(−2*,*3) split the line joining the original two points into four equal parts.

Let be the midpoint of *A*(−4*,*0) and *M*(−2*,*3).

Now let bet the midpoint of *B*(0*,*6) and *M*(−2*,*3).

Hence the co-ordinates of the points which divide the line joining the two given points are.

#### Page No 6.30:

#### Question 31:

Show that the mid-point of the line segment joining the points (5, 7) and (3, 9) is also the mid-point of the line segment joining the points (8, 6) and (0, 10).

#### Answer:

We have two points A (5, 7) and B (3, 9) which form a line segment and similarly

C (8, 6) and D (0, 10) form another line segment.

We have to prove that mid-point of AB is also the mid-point of CD.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of line segment AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is (4, 8)

Similarly mid-point Q of side CD can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (4, 8)

Hence the point P and Q coincides.

Thus mid-point of AB is also the mid-point of CD.

#### Page No 6.30:

#### Question 32:

Find the distance of the point (1, 2) from the mid-point of the line segment joining the points (6, 8) and (2, 4).

#### Answer:

We have to find the distance of a point A (1, 2) from the mid-point of the line segment joining P (6, 8) and Q (2, 4).

In general to find the mid-point of any two pointsand we use section formula as,

Therefore mid-point B of line segment PQ can be written as,

Now equate the individual terms to get,

So co-ordinates of B is (4, 6)

Therefore distance between A and B,

#### Page No 6.30:

#### Question 33:

If *A* and *B* are (1, 4) and (5, 2) respectively, find the coordinates of *P* when *AP*/*BP* = 3/4.

#### Answer:

The co-ordinates of the point dividing two points and in the ratio is given as,

where,

Here the two given points are *A*(1*,*4) and *B*(5*,*2). Let point *P*(*x, y*) divide the line joining ‘*AB*’ in the ratio

Substituting these values in the earlier mentioned formula we have,

Thus the co-ordinates of the point which divides the given points in the required ratio are.

#### Page No 6.30:

#### Question 34:

Show that the points *A* (1, 0), *B* (5, 3), *C* (2, 7) and *D* (−2, 4) are the vertices of a parallelogram.

#### Answer:

Let A (1, 0); B (5, 3); C (2, 7) and D (−2, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

So the mid-point of the diagonal AC is,

Similarly mid-point of diagonal BD is,

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

#### Page No 6.30:

#### Question 35:

Determine the ratio in which the point P (m, 6) divides the join of *A*(−4, 3) and *B*(2, 8). Also, find the value of *m*.

#### Answer:

Here we are given that the point *P*(*m,*6) divides the line joining the points *A*(−4*,*3) and *B*(2*,*8) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘*m*’.

Equating the individual components we have

Thus the value of ‘*m*’ is.

#### Page No 6.30:

#### Question 36:

Determine the ratio in which the point (−6, *a*) divides the join of *A* (−3, 1) and *B* (−8, 9). Also find the value of *a*.

#### Answer:

Here we are given that the point *P*(−6*,a*) divides the line joining the points *A*(−3*,*1) and *B*(−8*,*9) in some ratio.

Let us substitute these values in the earlier mentioned formula.

Equating the individual components we have

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘*a*’.

Equating the individual components we have

Thus the value of ‘*a*’ is.

#### Page No 6.30:

#### Question 37:

*ABCD* is a rectangle formed by joining the points *A* (−1, −1), *B*(−1 4) *C* (5 4) and *D* (5, −1). *P*, *Q*, *R* and *S* are the mid-points of sides *AB*, *BC*, *CD* and *DA* respectively. Is the quadrilateral *PQRS* a square? a rectangle? or a rhombus? Justify your answer.

#### Answer:

We have a rectangle ABCD formed by joining the points A (−1,−1); B (−1, 4); C (5, 4) and D (5,−1). The mid-points of the sides AB, BC, CD and DA are P, Q, R, S respectively.

We have to find that whether PQRS is a square, rectangle or rhombus.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point P of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of P is

Similarly mid-point Q of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of Q is (2, 4)

Similarly mid-point R of side CD can be written as,

Now equate the individual terms to get,

So co-ordinates of R is

Similarly mid-point S of side DA can be written as,

Now equate the individual terms to get,

So co-ordinates of S is (2,−1)

So we should find the lengths of sides of quadrilateral PQRS.

All the sides of quadrilateral are equal.

So now we will check the lengths of the diagonals.

All the sides are equal but the diagonals are unequal. Hence ABCD is a rhombus.

#### Page No 6.30:

#### Question 38:

Points P, Q, R and S divides the line segment joining A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q and R. [CBSE 2014]

#### Answer:

It is given that P, Q, R and S divides the line segment joining A(1, 2) and B(6, 7) in 5 equal parts.

∴ AP = PQ = QR = RS = SB .....(1)

Now,

AP + PQ + QR + RS + SB = AB

⇒ AP + AP + AP + AP + AP = AB [From (1)]

⇒ 5AP = AB

⇒ AP = $\frac{1}{5}$AB .....(2)

Now,

PB = PQ + QR + RS + SB = $\frac{1}{5}$AB +$\frac{1}{5}$AB + $\frac{1}{5}$AB + $\frac{1}{5}$AB = $\frac{4}{5}$AB .....(3)

From (2) and (3), we get

AP : PB = $\frac{1}{5}$AB : $\frac{4}{5}$AB = 1 : 4

Similarly,

AQ : QB = 2 : 3 and AR : RB = 3 : 2

Using section formula, we get

Coordinates of P = $\left(\frac{1\times 6+4\times 1}{1+4},\frac{1\times 7+4\times 2}{1+4}\right)=\left(\frac{10}{5},\frac{15}{5}\right)=\left(2,3\right)$

Coordinates of Q = $\left(\frac{2\times 6+3\times 1}{2+3},\frac{2\times 7+3\times 2}{2+3}\right)=\left(\frac{15}{5},\frac{20}{5}\right)=\left(3,4\right)$

Coordinates of R = $\left(\frac{3\times 6+2\times 1}{3+2},\frac{3\times 7+2\times 2}{3+2}\right)=\left(\frac{20}{5},\frac{25}{5}\right)=\left(4,5\right)$

#### Page No 6.30:

#### Question 39:

If *A* and *B* are two points having coordinates (−2, −2) and (2, −4) respectively, find the coordinates of *P* such that *AP* = $\frac{3}{7}$*AB.*

#### Answer:

We have two points A (−2,−2) and B (2,−4). Let P be any point which divide AB as,

Since,

So,

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

Therefore P divides AB in the ratio 3: 4. So,

#### Page No 6.30:

#### Question 40:

Find the coordinates of the points which divide the line segment joining A(−2, 2) and B (2, 8) into four equal parts.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Here we are supposed to find the points which divide the line joining *A*(−2*,*2) and *B*(2*,*8) into 4 equal parts.

We shall first find the midpoint *M*(*x, y*)* *of these two points since this point will divide the line into two equal parts.

So the point *M*(0*,*5) splits this line into two equal parts.

Now, we need to find the midpoint of *A*(−2*,*2) and *M*(0*,*5) separately and the midpoint of *B*(2*,*8) and *M*(0*,*5). These two points along with *M*(0*,*5) split the line joining the original two points into four equal parts.

Let be the midpoint of *A*(−2*,*2) and *M*(0*,*5).

Now let bet the midpoint of *B*(2*,*8) and *M*(0*,*5).

Hence the co-ordinates of the points which divide the line joining the two given points are.

#### Page No 6.30:

#### Question 41:

Three consecutive vertices of a parallelogram are (−2,−1), (1, 0) and (4, 3). Find the fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−2,−1); B (1, 0) and C (4, 3). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

#### Page No 6.30:

#### Question 42:

The points (3, −4) and (−6, 2) are the extremities of a diagonal of a parallelogram. If the third vertex is (−1,−3). Find the coordinates of the fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (3,−4); B (−1,−3) and C (−6, 2). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

#### Page No 6.30:

#### Question 43:

If the coordinates of the mid-points of the sides of a triangle are (1, 1) (2, −3) and (3, 4), find the vertices of the triangle.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be*,* and.

The three midpoints are given. Let these points be*, * and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘*x*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘*y*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

#### Page No 6.30:

#### Question 44:

Determine the ratio in which the straight line x − y − 2 = 0 divides the line segment joining (3, −1) and (8, 9).

#### Answer:

Let the line divide the line segment joining the points A (3,−1) and B (8, 9) in the ratio at any point

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

So,

Since, P lies on the given line. So,

Put the values of co-ordinates of point P in the equation of line to get,

On further simplification we get,

So,

So the line divides the line segment joining A and B in the ratio 2: 3 internally.

#### Page No 6.30:

#### Question 45:

Three vertices of a parallelogram are (a+b, a−b), (2a+b, 2a−b), (a−b, a+b). Find the fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are;and. We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the forth vertex is

#### Page No 6.30:

#### Question 46:

If two vertices of a parallelogram are (3, 2) (−1, 0) and the diagonals cut at (2, −5), find the other vertices of the parallelogram.

#### Answer:

We have a parallelogram ABCD in which A (3, 2) and B (−1, 0) and the co-ordinate of the intersection of diagonals is M (2,−5).

We have to find the co-ordinates of vertices C and D.

So let the co-ordinates be and

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the co-ordinate of vertex C is (1,−12)

Similarly,

Therefore,

Now equate the individual terms to get the unknown value. So,

So the co-ordinate of vertex C is (5,−10)

#### Page No 6.30:

#### Question 47:

If the coordinates of the mid-points of the sides of a triangle are (3, 4) (4, 6) and (5, 7), find its vertices.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be*,* and.

The three midpoints are given. Let these points be*, * and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘*x*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘*y*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

#### Page No 6.30:

#### Question 48:

The line segment joining the points *P*(3, 3) and *Q*(6, −6) is trisected at the points *A* and *B* such that *A* is nearer to *P*. If A also lies on the line given by 2*x* + *y* + *k* = 0, find the value of* k*.

#### Answer:

We have two points P (3, 3) and Q (6,−6). There are two points A and B which trisect the line segment joining P and Q.

Let the co-ordinate of A be

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point A is the point of trisection of the line segment PQ. So, A divides PQ in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point A is(4, 0)

It is given that point A lies on the line whose equation is

So point A will satisfy this equation.

So,

#### Page No 6.30:

#### Question 49:

If three consecutive vertices of a parallelogram are (1, −2), (3, 6) and (5, 10), find its fourth vertex.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (1,−2);

B (3, 6) and C(5, 10). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Now to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

So the forth vertex is

#### Page No 6.30:

#### Question 50:

(i) If the points A (*a*, −11), B (5, *b*), C(2, 15) and D (1, 1) are the vertices of a parallelogram *ABCD*, find the values of *a* and *b*.

(ii) Point A(3, 1), B(5, 1), C(*a, b*) and D(4, 3) are vertices of a parallelogram *ABCD*. Find the values of *a* and *b.*

#### Answer:

(i)

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (*a*,−11); B (5, *b*); C (2, 15) and D (1, 1).

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

(ii) Given: Point*A*(3, 1),

*B*(5, 1),

*C*(

*a, b*) and

*D*(4, 3) are vertices of a parallelogram

*ABCD*.

Diagonals of a parallelogram bisect each other.

∴ Mid point of

*AC*= Mid point of

*BD*

$\mathrm{Mid}\mathrm{point}\mathrm{of}\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right).\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Mid}\mathrm{point}\mathrm{of}AC=\left(\frac{3+a}{2},\frac{1+b}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Mid}\mathrm{point}\mathrm{of}BD=\left(\frac{5+4}{2},\frac{1+3}{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{9}{2},\frac{4}{2}\right)=\left(\frac{9}{2},2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \left(\frac{3+a}{2},\frac{1+b}{2}\right)=\left(\frac{9}{2},2\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3+a}{2}=\frac{9}{2}\mathrm{and}\frac{1+b}{2}=2\phantom{\rule{0ex}{0ex}}\Rightarrow 3+a=9\mathrm{and}1+b=4\phantom{\rule{0ex}{0ex}}\Rightarrow a=6\mathrm{and}b=3$

Hence, the values of

*a*and

*b*is 6 and 3, respectively.

#### Page No 6.31:

#### Question 51:

If the coordinates of the mid-points of the sides of a triangle be (3, −2), (−3, 1) and (4, −3), then find the coordinates of its vertices.

#### Answer:

The co-ordinates of the midpoint between two points and is given by,

Let the three vertices of the triangle be*,* and.

The three midpoints are given. Let these points be*, * and.

Let us now equate these points using the earlier mentioned formula,

Equating the individual components we get,

Using the midpoint of another side we have,

Equating the individual components we get,

Using the midpoint of the last side we have,

Equating the individual components we get,

Adding up all the three equations which have variable ‘*x*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Adding up all the three equations which have variable ‘*y*’ alone we have,

Substituting in the above equation we have,

Therefore,

And

Therefore the co-ordinates of the three vertices of the triangle are.

#### Page No 6.31:

#### Question 52:

(i) Points *P* and *Q* trisect the line segment joining the points *A*(–2, 0) and *B*(0, 8) such that *P* is near to *A*. Find the coordinates of *P* and *Q*.

(ii) The line segment joining the points (3, −4) and (1, 2) is trisected at the points *P* and *Q*. If the coordinates of *P* and *Q* are (*p*, −2) and (5/3, *q*) respectively. Find the values of* p* and *q*.

#### Answer:

(ii) Let the points *A*(3, −4) and *B*(1, 2) is trisected at the points *P*(*p*, −2) and *Q*(5/3, *q*).

Thus, *AP* = *PQ* = *QB*

Therefore, *P* divides *AB* internally in the ratio 1 : 2.

Section formula: if the point (*x*, *y*) divides the line segment joining the points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) internally in the ratio *m *:* n*, then the coordinates (*x*, *y*) = $\left(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$

Therefore, using section formula, the coordinates of *P* are:

$\left(p,-2\right)=\left(\frac{1\left(1\right)+2\left(3\right)}{1+2},\frac{1\left(2\right)+2\left(-4\right)}{1+2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(p,-2\right)=\left(\frac{1+6}{3},\frac{2-8}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(p,-2\right)=\left(\frac{7}{3},\frac{-6}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(p,-2\right)=\left(\frac{7}{3},-2\right)\phantom{\rule{0ex}{0ex}}\Rightarrow p=\frac{7}{3}$

Also, *Q* divides *AB* internally in the ratio 2 : 1.

Therefore, using section formula, the coordinates of *Q* are:

$\left(\frac{5}{3},q\right)=\left(\frac{2\left(1\right)+1\left(3\right)}{2+1},\frac{2\left(2\right)+1\left(-4\right)}{2+1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{5}{3},q\right)=\left(\frac{2+3}{3},\frac{4-4}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{5}{3},q\right)=\left(\frac{5}{3},\frac{0}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{5}{3},q\right)=\left(\frac{5}{3},0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow q=0$

Hence, the values of *p* and *q *is $\frac{7}{3}$ and 0, respectively.

#### Page No 6.31:

#### Question 53:

The line joining the points (2, 1) and (5, −8) is trisected at the points *P* and *Q*. If point *P* lies on the line 2*x* − *y* + *k* = 0. Find the value of k.

#### Answer:

We have two points A (2, 1) and B (5,−8). There are two points P and Q which trisect the line segment joining A and B.

Now according to the section formula if any point P divides a line segment joining andin the ratio m: n internally than,

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Therefore, co-ordinates of point P is(3,−2)

It is given that point P lies on the line whose equation is

So point A will satisfy this equation.

So,

#### Page No 6.31:

#### Question 54:

*A* (4, 2), *B*(6, 5) and *C* (1, 4) are the vertices of Δ*ABC*.

(i) The median from *A* meets *BC* in *D.* Find the coordinates of the point *D*.

(ii) Find the coordinates of point *P* and *AD* such that *AP* : *PD* = 2 : 1.

(iii) Find the coordinates of the points* Q* and *R* on medians *BE* and *CF *respectively such that *BQ* : *QE* = 2 : 1 and *CR* : *RF* = 2 : 1

(iv) What do you observe?

#### Answer:

We have triangle in which the co-ordinates of the vertices are A (4, 2); B (6, 5) and C (1, 4)

(i)It is given that median from vertex A meets BC at D. So, D is the mid-point of side BC.

In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point D of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of D is

(ii)We have to find the co-ordinates of a point P which divides AD in the ratio 2: 1 internally.

P divides AD in the ratio 2: 1. So,

(iii)We need to find the mid-point of sides AB and AC. Let the mid-points be F and E for the sides AB and AC respectively.

Therefore mid-point F of side AB can be written as,

So co-ordinates of F is

Similarly mid-point E of side AC can be written as,

So co-ordinates of E is

Q divides BE in the ratio 2: 1. So,

Similarly, R divides CF in the ratio 2: 1. So,

(iv)We observe that that the point P, Q and R coincides with the centroid. This also shows that centroid divides the median in the ratio 2: 1.

#### Page No 6.31:

#### Question 55:

If the points *A* (6, 1), *B* (8, 2), *C* (9, 4) and *D* (*k*, *p*) are the vertices of a parallelogram taken in order, then find the values of *k* and *p*.

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (6, 1); B (8, 2); C (9, 4) and D (*k, p*).

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

#### Page No 6.31:

#### Question 56:

A point P divides the line segment joining the points A(3, −5) and B(−4, 8) such that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$ If P lies on the line *x* + *y* = 0, then find the value of *k*. [CBSE 2012]

#### Answer:

It is given that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{k}{1}.$

So, P divides the line segment joining the points A(3, −5) and B(−4, 8) in the ratio *k* : 1.

Using the section formula, we get

Coordinates of P = $\left(\frac{-4k+3}{k+1},\frac{8k-5}{k+1}\right)$

Since P lies on the line *x* + *y* = 0, so

$\frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-4k+3+8k-5}{k+1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4k-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{1}{2}$

Hence, the value of *k* is $\frac{1}{2}$.

#### Page No 6.31:

#### Question 57:

The mid-point P of the line segment joining the points A(−10, 4) and B(−2, 0) lies on the line segment joining the points C(−9, −4) and D(−4, *y*). Find the ratio in which P divides CD. Also, find the value of *y*. [CBSE 2014]

#### Answer:

It is given that P is the mid-point of the line segment joining the points A(−10, 4) and B(−2, 0).

∴ Coordinates of P = $\left(\frac{-10+\left(-2\right)}{2},\frac{4+0}{2}\right)=\left(\frac{-12}{2},\frac{4}{2}\right)=\left(-6,2\right)$

Suppose P divides the line segment joining the points C(−9, −4) and D(−4, *y*) in the ratio *k* : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{-4k-9}{k+1},\frac{ky-4}{k+1}\right)$

$\therefore \left(\frac{-4k-9}{k+1},\frac{ky-4}{k+1}\right)=\left(-6,2\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-4k-9}{k+1}=-6\mathrm{and}\frac{ky-4}{k+1}=2$

Now,

$\frac{-4k-9}{k+1}=-6\phantom{\rule{0ex}{0ex}}\Rightarrow -4k-9=-6k-6\phantom{\rule{0ex}{0ex}}\Rightarrow 2k=3\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{3}{2}$

So, P divides the line segment CD in the ratio 3 : 2.

Putting *k* = $\frac{3}{2}$ in $\frac{ky-4}{k+1}=2$, we get

$\frac{{\displaystyle \frac{3y}{2}}-4}{{\displaystyle \frac{3}{2}}+1}=2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3y-8}{5}=2\phantom{\rule{0ex}{0ex}}\Rightarrow 3y-8=10\phantom{\rule{0ex}{0ex}}\Rightarrow 3y=18\phantom{\rule{0ex}{0ex}}\Rightarrow y=6$

Hence, the value of *y* is 6.

#### Page No 6.31:

#### Question 58:

If the point $C\left(-1,2\right)$ divides internally the line segment joining the points *A* (2, 5) and *B*( *x*, *y *) in the ratio 3 : 4 , find the value of *x*^{2} + *y*^{2} .

#### Answer:

It is given that the point C(–1, 2) divides the line segment joining the points A(2, 5) and B(*x*, *y*) in the ratio 3 : 4 internally.

Using the section formula, we get

$\left(-1,2\right)=\left(\frac{3\times x+4\times 2}{3+4},\frac{3\times y+4\times 5}{3+4}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(-1,2\right)=\left(\frac{3x+8}{7},\frac{3y+20}{7}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x+8}{7}=-1\mathrm{and}\frac{3y+20}{7}=2$

⇒ 3*x* + 8 = –7 and 3*y* + 20 = 14

⇒ 3*x* = –15 and 3*y* = –6

⇒ *x* = –5 and *y* = –2

∴ *x*^{2} + *y*^{2}^{ }= 25 + 4 = 29

Hence, the value of *x*^{2} + *y*^{2}^{ }is 29.

#### Page No 6.31:

#### Question 59:

* ABCD* is a parallelogram with vertices $A({x}_{1},{y}_{1}),B\left({x}_{2},{y}_{2}\right),C({x}_{3},{y}_{3})$ . Find the coordinates of the fourth vertex* D* in terms of ${x}_{1},{x}_{2},{x}_{3},{y}_{1},{y}_{2}\mathrm{and}{y}_{3}$.

#### Answer:

Suppose the coordinates of D be (*x*, *y*).

Since diagonals of a parallelogram bisect each other.

Therefore the midpoint of AC is the midpoint of BD, i.e $\left(\frac{{x}_{1}+{x}_{3}}{2},\frac{{\displaystyle {y}_{1}+{y}_{3}}}{{\displaystyle 2}}\right)and\left(\frac{{\displaystyle {x}_{2}+x}}{{\displaystyle 2}},\frac{{\displaystyle {y}_{2}+y}}{{\displaystyle 2}}\right)$ respectively.

$\Rightarrow {x}_{1}+{x}_{3}={x}_{2}+xand{y}_{1}+{y}_{3}={y}_{2}+y\phantom{\rule{0ex}{0ex}}\Rightarrow x={x}_{1}+{x}_{3}-{x}_{2}andy={y}_{1}+{y}_{3}-{y}_{2}\phantom{\rule{0ex}{0ex}}ThuscoordinatesofDare\left({x}_{1}+{x}_{3}-{x}_{2},{y}_{1}+{y}_{3}-{y}_{2}\right)$

#### Page No 6.31:

#### Question 60:

The points $A\left({x}_{1},{y}_{1}\right),B\left({x}_{2},{y}_{2}\right),C\left({x}_{3},{y}_{3}\right)$ are the vertices of $\u2206$ *ABC* .

(i) The median from *A *meets *BC *at *D* . Find the coordinates of the point *D*.

(ii) Find the coordinates of the point *P *on *AD* such that *AP* :* PD* = 2 : 1.

(iii) Find the points of coordinates* Q* and *R *on medians *BE *and *CF* respectively such that* BQ* : *QE* = 2 : 1 and *CR* : *RF *= 2 : 1.

(iv) What are the coordinates of the centropid of the triangle *ABC *?

#### Answer:

(i) Median AD of the triangle will divide the side BC in two equal parts.

Therefore, D is the midpoint of side BC.

Coordinates of D are

$\left(\frac{{x}_{2}+{x}_{3}}{2},\frac{{y}_{2}+{y}_{3}}{2}\right)$

(ii)

THe point P divided the side AD in the ratio 2: 1.

Coordinates of P are

$\left(\frac{2\times {\displaystyle \left(\frac{{x}_{2}+{x}_{3}}{2}\right)}+1\times {x}_{1}}{2+1},\frac{{\displaystyle 2\times \left(\frac{{y}_{2}+{y}_{3}}{2}\right)+1\times {y}_{1}}}{2+1}\right)=\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$

(iii)

Median BE of the triangle will divide the side AC in two equal parts.

Therefore, E is the midpoint of side AC.

Coordinates of E are

$\left(\frac{{x}_{1}+{x}_{3}}{2},\frac{{y}_{1}+{y}_{3}}{2}\right)$

The point Q divided the side BE in the ratio 2: 1.

Coordinates of Q are

$\left(\frac{2\times {\displaystyle \left(\frac{{x}_{1}+{x}_{3}}{2}\right)}+1\times {x}_{2}}{2+1},\frac{{\displaystyle 2\times \left(\frac{{y}_{1}+{y}_{3}}{2}\right)+1\times {y}_{2}}}{2+1}\right)=\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$

Similarly, Coordinates of Q are R are $\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$.

(iv)

The points P, Q and R coincides and is the centroid of the triangle ABC.

So, coordinates of the centroid is $\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$.

#### Page No 6.37:

#### Question 1:

Find the centroid of the triangle whose vertices are:

(i) (1, 4) (−1,−1), (3, −2)

(ii) (−2, 3) (2, −1) (4, 0)

#### Answer:

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

(i) The co-ordinates of the centroid of a triangle whose vertices are (1, 4); (−1,−1); (3,−2) are-

(ii) The co-ordinates of the centroid of a triangle whose vertices are (−2, 3); (2,−1); (4, 0) are-

#### Page No 6.37:

#### Question 2:

Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the coordinates of the third vertex.

#### Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are (1, 2) and (3, 5)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is−

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 6.37:

#### Question 3:

Find the third vertex of a triangle, if two of its vertices are at (−3, 1) and (0, −2) and the centroid is at the origin.

#### Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are (−3, 1) and (0, −2)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 6.37:

#### Question 4:

*A* (3, 2) and *B* (−2, 1) are two vertices of a triangle *ABC *whose centroid *G* has the coordinates $\left(\frac{5}{3},-\frac{1}{3}\right)$. Find the coordinates of the third vertex *C* of the triangle.

#### Answer:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be.

The co-ordinates of other two vertices are A (3, 2) and C (−2, 1)

The co-ordinate of the centroid is

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 6.37:

#### Question 5:

If (−2, 3), (4, −3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid.

#### Answer:

Letbe ant triangle such that P (−2, 3); Q (4,−3) and R (4, 5) are the mid-points of the sides AB, BC, CA respectively.

We have to find the co-ordinates of the centroid of the triangle.

Let the vertices of the triangle be

In general to find the mid-point of two pointsand we use section formula as,

So, co-ordinates of P,

Equate the *x* component on both the sides to get,

…… (1)

Similarly,

…… (2)

Similarly, co-ordinates of Q,

Equate the *x* component on both the sides to get,

…… (3)

Similarly,

…… (4)

Similarly, co-ordinates of R,

Equate the *x* component on both the sides to get,

…… (5)

Similarly,

…… (6)

Add equation (1) (3) and (5) to get,

Similarly, add equation (2) (4) and (6) to get,

We know that the co-ordinates of the centroid G of a triangle whose vertices are is-

So, centroid G of a triangle is,

#### Page No 6.37:

#### Question 6:

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

#### Answer:

Letbe any triangle such that O is the origin and the other co-ordinates are. P and R are the mid-points of the sides OA and OB respectively.

We have to prove that line joining the mid-point of any two sides of a triangle is equal to half of the third side which means,

In general to find the mid-point of two pointsand we use section formula as,

So,

Co-ordinates of P is,

Similarly, co-ordinates of R is,

In general, the distance between A and B is given by,

Similarly,

Hence,

#### Page No 6.38:

#### Question 7:

Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.

#### Answer:

Let us consider a Cartesian plane having a parallelogram OABC in which O is the origin.

We have to prove that middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.

Let the co-ordinate of A be. So the coordinates of other vertices of the quadrilateral are- O (0, 0); B; C

Let P, Q, R and S be the mid-points of the sides AB, BC, CD, DA respectively.

In general to find the mid-point of two pointsand we use section formula as,

So co-ordinate of point P,

Similarly co-ordinate of point Q,

Similarly co-ordinate of point R,

Similarly co-ordinate of point S,

Let us find the co-ordinates of mid-point of PR as,

Similarly co-ordinates of mid-point of QS as,

Now the mid-point of diagonal AC,

Similarly the mid−point of diagonal OA,

Hence the mid-points of PR, QS, AC and OA coincide.

Thus, middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.

#### Page No 6.38:

#### Question 8:

If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA^{2} + PB^{2} + PC^{2} = GA^{2} + GB^{2} + GC^{2} + 3GP^{2}.

#### Answer:

Letbe any triangle whose coordinates are. Let P be the origin and G be the centroid of the triangle.

We have to prove that,

…… (1)

We know that the co-ordinates of the centroid G of a triangle whose vertices are is−

In general, the distance between A and B is given by,

So,

Now,

So we get the value of left hand side of equation (1) as,

Similarly we get the value of right hand side of equation (1) as,

Hence,

#### Page No 6.38:

#### Question 9:

If *G* be the centroid of a triangle ABC, prove that:

AB^{2} + BC^{2} + CA^{2} = 3 (GA^{2} + GB^{2} + GC^{2})

#### Answer:

$\mathrm{Let}\mathrm{A}\left({x}_{1},{y}_{1}\right);\mathrm{B}\left({x}_{2},{y}_{2}\right);\mathrm{C}\left({x}_{3},{y}_{3}\right)\mathrm{be}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{vertices}\mathrm{of}\u2206\mathrm{ABC}.\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{us}\mathrm{assume}\mathrm{that}\mathrm{centroid}\mathrm{of}\mathrm{the}\u2206\mathrm{ABC}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{origin}\mathrm{G}.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{G}\mathrm{are}\mathrm{G}\left(0,0\right).\phantom{\rule{0ex}{0ex}}\mathrm{Now},\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3}=0;\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}=0\phantom{\rule{0ex}{0ex}}\mathrm{so},{x}_{1}+{x}_{2}+{x}_{3}=0.......\left(1\right)\phantom{\rule{0ex}{0ex}}{y}_{1}+{y}_{2}+{y}_{3}=0........\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+2{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+2{x}_{3}{x}_{1}=0.......\left(3\right)\phantom{\rule{0ex}{0ex}}{{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}+2{y}_{1}{y}_{2}+2{y}_{2}{y}_{3}+2{y}_{3}{y}_{1}=0......\left(4\right)\phantom{\rule{0ex}{0ex}}\mathrm{LHS}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}+{\mathrm{CA}}^{2}\phantom{\rule{0ex}{0ex}}={\left[\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\right]}^{2}+{\left[\sqrt{{\left({x}_{3}-{x}_{2}\right)}^{2}+{\left({y}_{3}-{y}_{2}\right)}^{2}}\right]}^{2}+{\left[\sqrt{{\left({x}_{3}-{x}_{1}\right)}^{2}+{\left({y}_{3}-{y}_{1}\right)}^{2}}\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({x}_{3}-{x}_{2}\right)}^{2}+{\left({y}_{3}-{y}_{2}\right)}^{2}+{\left({x}_{3}-{x}_{1}\right)}^{2}+{\left({y}_{3}-{y}_{1}\right)}^{2}\phantom{\rule{0ex}{0ex}}={{x}_{1}}^{2}+{{x}_{2}}^{2}-2{x}_{1}{x}_{2}+{{y}_{1}}^{2}+{{y}_{2}}^{2}-2{y}_{1}{y}_{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}-2{x}_{2}{x}_{3}+{{y}_{2}}^{2}+{{y}_{3}}^{2}-2{y}_{2}{y}_{3}+{{x}_{1}}^{2}+{{x}_{3}}^{2}-2{x}_{1}{x}_{3}+{{y}_{1}}^{2}+{{y}_{3}}^{2}-2{y}_{1}{y}_{3}\phantom{\rule{0ex}{0ex}}=2\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}\right)+2\left({{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)-\left(2{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+2{x}_{3}{x}_{1}\right)-\left(2{y}_{1}{y}_{2}+2{y}_{2}{y}_{3}+2{y}_{3}{y}_{1}\right)\phantom{\rule{0ex}{0ex}}=2\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}\right)+2\left({{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)+\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}\right)+\left({{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)\phantom{\rule{0ex}{0ex}}=3\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{RHS}=3\left({\mathrm{GA}}^{2}+{\mathrm{GB}}^{2}+{\mathrm{GC}}^{2}\right)\phantom{\rule{0ex}{0ex}}=3\left[{\left\{\sqrt{{\left({x}_{1}-0\right)}^{2}+{\left({y}_{1}-0\right)}^{2}}\right\}}^{2}+{\left\{\sqrt{{\left({x}_{2}-0\right)}^{2}+{\left({y}_{2}-0\right)}^{2}}\right\}}^{2}+{\left\{\sqrt{{\left({x}_{3}-0\right)}^{2}+{\left({y}_{3}-0\right)}^{2}}\right\}}^{2}\right]\phantom{\rule{0ex}{0ex}}=3\left({{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{y}_{1}}^{2}+{{y}_{2}}^{2}+{{y}_{3}}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}+{\mathrm{CA}}^{2}=3\left({\mathrm{GA}}^{2}+{\mathrm{GB}}^{2}+{\mathrm{GC}}^{2}\right)$

#### Page No 6.38:

#### Question 10:

In Fig. 14.36, a right triangle BOA is given C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.

#### Answer:

We have a right angled triangle, right angled at O. Co-ordinates are B (0,2*b*); A (2*a**, *0) and C (0, 0).

We have to prove that mid-point C of hypotenuse AB is equidistant from the vertices.

In general to find the mid-point of two pointsand we use section formula as,

So co-ordinates of C is,

In general, the distance between A and B is given by,

So,

Hence, mid−point C of hypotenuse AB is equidistant from the vertices.

#### Page No 6.4:

#### Question 1:

On which axis do the following points lie?

(a) P(5, 0)

(b) Q(0−2)

(c) R(−4,0)

(d) S(0,5)

#### Answer:

According to the Rectangular Cartesian Co-ordinate system of representing a point *(x, y)*,

If then the point lies in the 1^{st} quadrant

If then the point lies in the 2^{nd} quadrant

If then the point lies in the 3^{rd} quadrant

If then the point lies in the 4^{th} quadrant

But in case

If then the point lies on the *y*-axis

If then the point lies on the *x*-axis

(i) Here the point is given to be *P *(5, 0). Comparing this with the standard form of

(*x, y*) we have

Here we see that

Hence the given point lies on the

(ii) Here the point is given to be *Q* *(0, *-−2). Comparing this with the standard form of *(x, y)* we have

Here we see that

Hence the given point lies on the

(iii) Here the point is given to be *R *(-4, 0). Comparing this with the standard form of (*x, y*) we have

Here we see that

Hence the given point lies on the

(iv) Here the point is given to be S (0, 5). Comparing this with the standard form of (*x, y*) we have

Here we see that

Hence the given point lies on the

#### Page No 6.4:

#### Question 2:

Let *ABCD* be a square of side 2*a*. Find the coordinates of the vertices of this square when

(i) A coincides with the origin and *AB* and *AD* are along *OX* and *OY* respectively.

(ii) The centre of the square is at the origin and coordinate axes are parallel to the sides *AB* and *AD* respectively.

#### Answer:

The distance between any two adjacent vertices of a square will always be equal. This distance is nothing but the side of the square.

Here, the side of the square ‘*ABCD*’ is given to be ‘2*a*’.

(i) Since it is given that the vertex ‘*A*’ coincides with the origin we know that the co-ordinates of this point is (0, 0).

We also understand that the side ‘*AB*’ is along the *x*-axis. So, the vertex ‘*B*’ has got to be at a distance of ‘2*a*’ from ‘*A*’.

Hence the vertex ‘*B*’ has the co-ordinates (2*a**, *0).

Also it is said that the side ‘*AD*’ is along the *y*-axis. So, the vertex ‘*D*’ it has got to be at a distance of ‘*2a*’ from ‘*A*’.

Hence the vertex ‘*D*’ has the co-ordinates (0, 2*a*)

Finally we have vertex ‘*C*’ at a distance of ‘2*a*’ both from vertex ‘*B*’ as well as ‘*D*’.

Hence the vertex of ‘*C*’ has the co-ordinates (2*a**, *2*a*)

So, the co-ordinates of the different vertices of the square are

(ii) Here it is said that the centre of the square is at the origin and that the sides of the square are parallel to the axes.

Moving a distance of half the side of the square in either the ‘*upward*’ or ‘*downward*’ direction and also along either the ‘*right*’ or ‘*left*’ direction will give us all the four vertices of the square.

Half the side of the given square is ‘*a*’.

The centre of the square is the origin and its vertices are (0, 0).* *Moving a distance of ‘*a*’ to the right as well as up will lead us to the vertex ‘*A*’ and it will have vertices (*a, a*).

Moving a distance of ‘*a*’ to the left as well as up will lead us to the vertex ‘*B*’ and it will have vertices (-(−*a**, a*).

Moving a distance of ‘*a*’ to the left as well as down will lead us to the vertex ‘*C*’ and it will have vertices (-(−*a**, *-−*a*).

Moving a distance of ‘*a*’ to the right as well as down will lead us to the vertex ‘*D*’ and it will have vertices (*a,-,−a*).

So, the co-ordinates of the different vertices of the square are

#### Page No 6.4:

#### Question 3:

The base *PQ* of two equilateral triangles *PQR* and *PQR'* with side 2*a* lies along y-axis such that the mid-point of *PQ* is at the origin. Find the coordinates of the vertices *R* and* R*' of the triangles.

#### Answer:

In an equilateral triangle, the height ‘*h*’ is given by

Here it is given that ‘*PQ*’ forms the base of two equilateral triangles whose side measures ‘*2a*’ units.

The height of these two equilateral triangles has got to be

In an equilateral triangle the height drawn from one vertex meets the midpoint of the side opposite this vertex.

So here we have ‘*PQ*’ being the base lying along the *y*-axis with its midpoint at the origin, that is at *(0, 0)*.

So the vertices ‘*R*’ and ‘*R’*’ will lie perpendicularly to the *y*-axis on either sides of the origin at a distance of ‘’ units.

Hence the co-ordinates of ‘*R*’ and ‘*R’*’ are

#### Page No 6.53:

#### Question 1:

Find the area of a triangle whose vertices are

(i) (6, 3) (−3, 5) and (4, −2)

(ii) $(a{t}_{1}^{2},2a{t}_{1}),(a{t}_{2}^{2},2a{t}_{2})and(a{t}_{3}^{2},2a{t}_{3})$

(iii) (*a, c + a*), *(a, c*) and (−*a, c − a*)

(iv) (1, –1), (–4, 6) and (–3, –5)

#### Answer:

We know area of triangle formed by three points is given by

(i) The vertices are given as (6, 3), (−3, 5), (4, −2).

(ii) The vertices are given as .

(iii)

The vertices are given as .

(iv)Area of the triangle formed by the vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}$

$\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$

Now, the given vertices are (1, –1), (–4, 6) and (–3, –5)

Therefore,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|1\left(6-\left(-5\right)\right)+\left(-4\right)\left(-5-\left(-1\right)\right)+\left(-3\right)\left(-1-6\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|1\left(6+5\right)+\left(-4\right)\left(-5+1\right)+\left(-3\right)\left(-7\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|11+\left(-4\right)\left(-4\right)+21\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|11+16+21\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|48\right|\phantom{\rule{0ex}{0ex}}=24$

Hence, the area of a triangle is 24 square units.

#### Page No 6.53:

#### Question 2:

Find the area of the quadrilaterals, the coordinates of whose vertices are

(i) (−3, 2), (5, 4), (7, −6) and (−5, −4)

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

(iii) (−4, −2, (−3, −5), (3, −2), (2, 3)

#### Answer:

(i)

Let the vertices of the quadrilateral be A (−3, 2), B (5, 4), C (7, −6), and D (−5, −4). Join AC to form two triangles ΔABC and ΔACD.

(ii)

Let the vertices of the quadrilateral be A (1, 2), B (6, 2), C (5, 3), and D (3, 4). Join AC to form two triangles ΔABC and ΔACD.

(iii)

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

#### Page No 6.53:

#### Question 3:

The four vertices of a quadrilateral are (1, 2), (−5, 6), (7, −4) and (k, −2) taken in order. If the area of the quadrilateral is zero, find the value of *k*.

#### Answer:

GIVEN: The four vertices of quadrilateral are (1, 2), (−5, 6), (7, −4) and D (*k*, −2) taken in order. If the area of the quadrilateral is zero

TO FIND: value of *k*

PROOF: Let four vertices of quadrilateral are A (1, 2) and B (−5, 6) and C (7, −4) and D (*k*, −2)

We know area of triangle formed by three points is given by

Now Area of ΔABC

Taking three points when A (1, 2) and B (−5, 6) and C (7, −4)

Also,

Now Area of ΔACD

Taking three points when A (1, 2) and C (7, −4) and D (*k*, −2)

Hence

#### Page No 6.53:

#### Question 4:

The vertices of Δ*ABC* are (−2, 1), (5, 4) and (2, −3) respectively. Find the area of the triangle and the length of the altitude through *A*.

#### Answer:

GIVEN: The vertices of triangle ABC are A (−2, 1) and B (5, 4) and C (2, −3)

TO FIND: The area of triangle ABC and length if the altitude through A

PROOF: We know area of triangle formed by three points is given by

Now Area of ΔABC

Taking three points A (−2, 1) and B (5, 4) and C(2, −3)

We have

Now,

#### Page No 6.53:

#### Question 5:

Show that the following sets of points are collinear.

(a) (2, 5), (4, 6) and (8, 8)

(b) (1, −1), (2, 1) and (4, 5)

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

We know area of triangle formed by three points is given by

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(2*,* 5)*, B*(4*,* 6) and *C*(8*,* 8). Substituting these values in the earlier mentioned formula we have,

$A=\frac{1}{2}\left|2\left(6-8\right)+4\left(8-5\right)+8\left(5-6\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|2\left(-2\right)+4\left(3\right)+8\left(-1\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-4+12-8\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-12+12\right|\phantom{\rule{0ex}{0ex}}=0$

Since the area enclosed by the three points is equal to 0, the three points need to be.

The three given points are *A*(1*,* −1)*, B*(2*,* 1) and *C*(4*,* 5). Substituting these values in the earlier mentioned formula we have,

$A=\frac{1}{2}\left|1\left(1-5\right)+2\left(5+1\right)+4\left(-1-1\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|1\left(-4\right)+2\left(6\right)+4\left(-2\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-4+12-8\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-12+12\right|\phantom{\rule{0ex}{0ex}}=0$

Since the area enclosed by the three points is equal to 0, the three points need to be.

#### Page No 6.53:

#### Question 6:

Q

#### Answer:

Let the vertices of the quadrilateral be A (−3, 2), B (5, 4), C (7, −6), and D (−5, −4). Join AC to form two triangles ΔABC and ΔACD.

#### Page No 6.54:

#### Question 7:

In $\u2206$*ABC* , the coordinates of vertex* A *are (0, $-$1) and *D* (1,0) and* E*(0,10) respectively the mid-points of the sides *AB *and *AC *. If *F* is the mid-points of the side *BC *, find the area of $\u2206$* DEF.*

#### Answer:

Let the coordinates of B and C be $\left({x}_{2},{y}_{2}\right)$ and $\left({x}_{3},{y}_{3}\right)$, respectively.

D is the midpoint of AB.

So,

$\left(1,0\right)=\left(\frac{{x}_{2}+0}{2},\frac{{y}_{2}-1}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{{x}_{2}}{2}\mathrm{and}0=\frac{{y}_{2}-1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{2}=2\mathrm{and}{y}_{2}=1$

Thus, the coordinates of B are (2, 1).

Similarly, E is the midpoint of AC.

So,

$\left(0,1\right)=\left(\frac{{x}_{3}+0}{2},\frac{{y}_{3}-1}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 0=\frac{{x}_{3}}{2}\mathrm{and}1=\frac{{y}_{3}-1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{3}=0\mathrm{and}{y}_{3}=3$

Thus, the coordinates of C are (0, 3).

Also, F is the midpoint of BC. So, its coordinates are

$\left(\frac{2+0}{2},\frac{1+3}{2}\right)=\left(1,2\right)$

Now,

Area of a triangle = $\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]$

Thus, the area of $\u2206$ABC is

$\frac{1}{2}\left[0\left(1-3\right)+2\left(3+1\right)+0\left(-1-1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 8\phantom{\rule{0ex}{0ex}}=4\mathrm{square}\mathrm{units}$

And the area of $\u2206$DEF is

$\frac{1}{2}\left[1\left(1-2\right)+0\left(2-0\right)+1\left(0-1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \left(-2\right)\phantom{\rule{0ex}{0ex}}=1\mathrm{square}\mathrm{unit}\left(\mathrm{Taking}\mathrm{the}\mathrm{numerical}\mathrm{value},\mathrm{as}\mathrm{the}\mathrm{area}\mathrm{cannot}\mathrm{be}\mathrm{negative}\right)$

#### Page No 6.54:

#### Question 8:

Find the area of the triangle PQR with Q(3, 2) and the mid-points of the sides through Q being (2, −1) and (1, 2). [CBSE 2015]

#### Answer:

Let P(*x*_{1}, *y*_{1}), Q(3, 2) and R(*x*_{2}, *y*_{2}) be the vertices of the ∆PQR.

Suppose S(2, −1) and T(1, 2) be the mid-points of sides QR and PQ, respectively.

Using mid-point formula, we have

$\frac{{x}_{1}+3}{2}=1,\frac{{y}_{1}+2}{2}=2\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}+3=2,{y}_{1}+2=4\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}=-1,{y}_{1}=2$

∴ Coordinates of P = (−1, 2)

Also,

$\frac{{x}_{2}+3}{2}=2,\frac{{y}_{2}+2}{2}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{2}+3=4,{y}_{2}+2=-2\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{2}=1,{y}_{2}=-4$

∴ Coordinates of R = (1, −4)

So, P(−1, 2), Q(3, 2) and R(1, −4) are the vertices of ∆PQR.

$\therefore \mathrm{ar}\left(\u2206\mathrm{PQR}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-1\left[2-\left(-4\right)\right]+3\left(-4-2\right)+1\left(2-2\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-6-18+0\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-24\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 24\phantom{\rule{0ex}{0ex}}=12\mathrm{square}\mathrm{units}$

Hence, the area of the triangle is 12 square units.

#### Page No 6.54:

#### Question 9:

If P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area. [CBSE 2015]

#### Answer:

It is given that P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS.

Area of the quadrilateral PQRS = Area of ∆PQR + Area of ∆PRS

$\mathrm{ar}\left(\u2206\mathrm{PQR}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-5\left[-6-\left(-3\right)\right]+\left(-4\right)\left[-3-\left(-3\right)\right]+2\left[-3-\left(-6\right)\right]\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|15+0+6\right|\phantom{\rule{0ex}{0ex}}=\frac{21}{2}\mathrm{square}\mathrm{units}$

$\mathrm{ar}\left(\u2206\mathrm{PRS}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-5\left(-3-2\right)+2\left[2-\left(-3\right)\right]+1\left[-3-\left(-3\right)\right]\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|25+10+0\right|\phantom{\rule{0ex}{0ex}}=\frac{35}{2}\mathrm{square}\mathrm{units}$

∴ Area of the quadrilateral PQRS = $\frac{21}{2}+\frac{35}{2}=\frac{56}{2}=28$ square units

Hence, the area of the given quarilateral is 28 square units.

#### Page No 6.54:

#### Question 10:

If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area. [CBSE 2014]

#### Answer:

It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

$\mathrm{ar}\left(\u2206\mathrm{ABC}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-3\left[-7-\left(-8\right)\right]+\left(-2\right)\left(-8-5\right)+1\left[5-\left(-7\right)\right]\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-3+26+12\right|\phantom{\rule{0ex}{0ex}}=\frac{35}{2}\mathrm{square}\mathrm{units}$

$\mathrm{ar}\left(\u2206\mathrm{ACD}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-3\left(-8-3\right)+1\left(3-5\right)+6\left[5-\left(-8\right)\right]\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|33-2+78\right|\phantom{\rule{0ex}{0ex}}=\frac{109}{2}\mathrm{square}\mathrm{units}$

∴ Area of the quadrilateral ABCD = $\frac{35}{2}+\frac{109}{2}=\frac{144}{2}=72$ square units

Hence, the area of the given quadrilateral is 72 square units.

#### Page No 6.54:

#### Question 11:

For what value of *a* the point (*a*, 1), (1, −1), and (11, 4) are collinear?

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(*a,* 1)*, B*(1*, −*1) and *C*(11*,* 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(a\times -1+1\times 4+11\times 1\right)-\left(1\times 1+11\times -1+a\times 4\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-a+4+11\right)-\left(1-11+4a\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-a+15\right)-\left(-10+4a\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-a+15+10-4a\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-5a+25\right|\phantom{\rule{0ex}{0ex}}0=-5a+25\phantom{\rule{0ex}{0ex}}5a=25\phantom{\rule{0ex}{0ex}}a=5$

Hence the value of ‘*a*’ for which the given points are collinear is.

#### Page No 6.54:

#### Question 12:

Prove that the points (a, b), (a_{1}, b_{1}) and (a −a_{1}, b −b_{1}) are collinear if ab_{1} = a_{1}b.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are*,* and. If they are collinear then the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(a{b}_{1}+{a}_{1}\left(b-{b}_{1}\right)+\left(a-{a}_{1}\right)b\right)-\left({a}_{1}b+\left(a-{a}_{1}\right){b}_{1}+a\left(b-{b}_{1}\right)\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(a{b}_{1}+{a}_{1}b-{a}_{1}{b}_{1}+ab-{a}_{1}b\right)-\left({a}_{1}b+a{b}_{1}-{a}_{1}{b}_{1}+ab-a{b}_{1}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|a{b}_{1}+{a}_{1}b-{a}_{1}{b}_{1}+ab-{a}_{1}b-{a}_{1}b-a{b}_{1}+{a}_{1}{b}_{1}-ab+a{b}_{1}\right|\phantom{\rule{0ex}{0ex}}0=a{b}_{1}-{a}_{1}b\phantom{\rule{0ex}{0ex}}a{b}_{1}={a}_{1}b$

Hence we have proved that for the given conditions to be satisfied we need to have.

#### Page No 6.54:

#### Question 13:

(i) If the vertices of a triangle are (1, −3), (4, *p*) and (−9, 7) and its area is 15 sq. units, find the value(s) of *p*.

(ii) Find the value of *k* so that the area of triangle *ABC* with A*(k* + 1, 1), *B*(4, –3) and *C*(7, –*k*) is 6 square units.

#### Answer:

(i) Let A(1, −3), B(4, *p*) and C(−9, 7) be the vertices of the ∆ABC.

Here, *x*_{1} = 1, *y*_{1} = −3; *x*_{2} = 4, *y*_{2} = *p *and *x*_{3} = −9, *y*_{3} = 7

ar(∆ABC) = 15 square units

$\Rightarrow \frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=15\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|1\left(p-7\right)+4\left[7-\left(-3\right)\right]+\left(-9\right)\left(-3-p\right)\right|=15\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|p-7+40+27+9p\right|=15\phantom{\rule{0ex}{0ex}}\Rightarrow \left|10p+60\right|=30$

$\Rightarrow 10p+60=30$ or $10p+60=-30$

$\Rightarrow 10p=-30$ or $10p=-90$

$\Rightarrow p=-3$ or $p=-9$

Hence, the value of *p *is −3 or −9.

(ii) Area of the triangle formed by the vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}$ $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given vertices are A*(k* + 1, 1), *B*(4, –3) and *C*(7, –*k*)

and the given area is 6 square units.

Therefore,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|\left(k+1\right)\left(-3-\left(-k\right)\right)+\left(4\right)\left(-k-1\right)+\left(7\right)\left(1-\left(-3\right)\right)\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 6=\frac{1}{2}\left|\left(k+1\right)\left(-3+k\right)+\left(4\right)\left(-k-1\right)+\left(7\right)\left(1+3\right)\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 6=\frac{1}{2}\left|-3k+{k}^{2}-3+k-4k-4+\left(7\right)\left(4\right)\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 6=\frac{1}{2}\left|-3k+{k}^{2}-3+k-4k-4+28\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 12=\left|{k}^{2}-6k+21\right|\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}-6k+21=12\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}-6k+21-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}-6k+9=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(k-3\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow k-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=3$

Hence, the value of *k* is 3.

#### Page No 6.54:

#### Question 14:

If (x, y) be on the line joining the two points (1, −3) and (−4, 2) , prove that x + y + 2= 0.

#### Answer:

Since the point (*x*, *y*) lie on the line joining the points (1, −3) and (−4, 2); the area of triangle formed by these points is 0.

That is,

Thus, the result is proved.

#### Page No 6.54:

#### Question 15:

Find the value of* k* if points A*(k*, 3), B(6, −2) and C(−3, 4) are collinear.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(*k,* 3)*, B*(6*, −*2) and *C*(*−*3*,* 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(k\left(-2\right)+6\times 4+\left(-3\right)\times 3\right)-\left(6\times 3+\left(-3\right)\left(-2\right)+k\times 4\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-2k+24-9\right)-\left(18+6+4k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-2k+15-24-4k\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-6k-9\right|\phantom{\rule{0ex}{0ex}}0=-6k-9\phantom{\rule{0ex}{0ex}}k=-\frac{9}{6}=-\frac{3}{2}$

Hence the value of ‘*k*’ for which the given points are collinear is.

#### Page No 6.54:

#### Question 16:

Find the value of k, if the points *A*(7, −2), *B* (5, 1) and *C *(3, 2*k*) are collinear.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(7*, −*2)*, B*(5*,* 1) and *C*(3*,* 2*k*). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(7\times 1+5\times 2k+3\times -2\right)-\left(5\times -2+3\times 1+7\times 2k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(7+10k-6\right)-\left(-10+3+14k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-4k+8\right|\phantom{\rule{0ex}{0ex}}0=-4k+8\phantom{\rule{0ex}{0ex}}k=2$

Hence the value of ‘*k*’ for which the given points are collinear is.

#### Page No 6.54:

#### Question 17:

If the point P (*m*, 3) lies on the line segment joining the points $A\left(-\frac{2}{5},6\right)$and *B* (2, 8), find the value of *m*.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point *P*(*m,*3) lies on the line segment joining the points* * and *B*(2*,*8). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(-\frac{2}{5}\times 3+m\times 8+2\times 6\right)-\left(m\times 6+2\times 3+\left(-\frac{2}{5}\right)\times 8\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-\frac{6}{5}+8m+12\right)-\left(6m+6-\frac{16}{5}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|2m+8\right|\phantom{\rule{0ex}{0ex}}0=2m+8\phantom{\rule{0ex}{0ex}}m=-4$

Hence the value of ‘*m*’ for which the given condition is satisfied is.

#### Page No 6.54:

#### Question 18:

If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that *x *+ *y* = *a* + *b*.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point *R*(*x, y*) lies on the line segment joining the points *P*(*a, b*)* *and *Q*(*b, a*). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(ay+xa+{b}^{2}\right)-\left(xb+by+{a}^{2}\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|ay+xa+{b}^{2}-xb-by-{a}^{2}\right|\phantom{\rule{0ex}{0ex}}0=ay+xa+{b}^{2}-xb-by-{a}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}-{b}^{2}=ax+ay-bx-by\phantom{\rule{0ex}{0ex}}\left(a+b\right)\left(a-b\right)=\left(a-b\right)\left(x+y\right)\phantom{\rule{0ex}{0ex}}\left(a+b\right)=\left(x+y\right)\phantom{\rule{0ex}{0ex}}$

Hence under the given conditions we have proved that.

#### Page No 6.54:

#### Question 19:

Find the value of k, if the points A (8, 1) B(3, −4) and C(2, k) are collinear.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(8*,*1)*, B*(3*,**−*4) and *C*(2*,k*). It is also said that they are collinear and hence the area enclosed by them should be 0.

$\u2206=\frac{1}{2}\left|\left(8\times -4+3\times k+2\times 1\right)-\left(3\times 1+2\times -4+8\times k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|\left(-32+3k+2\right)-\left(3-8+8k\right)\right|\phantom{\rule{0ex}{0ex}}0=\frac{1}{2}\left|-25-5k\right|\phantom{\rule{0ex}{0ex}}k=-5$

Hence the value of ‘*k*’ for which the given points are collinear is.

#### Page No 6.54:

#### Question 20:

Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.

#### Answer:

*A*’ encompassed by three points, and is given by the formula,

$\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

The three given points are *A*(*a,*2*a*)*, B*(*−*2*,*6) and *C*(3*,*1). It is also said that the area enclosed by them is 10 square units. Substituting these values in the above mentioned formula we have,

$\u2206=\frac{1}{2}\left|\left(a\times 6+\left(-2\right)\times 1+3\times 2a\right)-\left(\left(-2\right)\times 2a+3\times 6+a\times 1\right)\right|\phantom{\rule{0ex}{0ex}}10=\frac{1}{2}\left|\left(6a-2+6a\right)-\left(-4a+18+a\right)\right|\phantom{\rule{0ex}{0ex}}10=\frac{1}{2}\left|15a-20\right|\phantom{\rule{0ex}{0ex}}20=\left|15a-20\right|\phantom{\rule{0ex}{0ex}}4=\left|3a-4\right|$

We have. Hence either

Or

Hence the values of ‘*a*’ which satisfies the given conditions are.

#### Page No 6.54:

#### Question 21:

If $a\ne b\ne 0$, prove that the points (*a*, *a*^{2}), (*b*, *b*^{2}) (0, 0) will not be collinear.

#### Answer:

Let A(*a*, *a*^{2}), B(*b*, *b*^{2}) and C(0, 0) be the coordinates of the given points.

We know that the area of triangle having vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)$ is $\left|\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\right|$ square units.

So,

Area of ∆ABC

$=\left|\frac{1}{2}\left[a\left({b}^{2}-0\right)+b\left(0-{a}^{2}\right)+0\left({a}^{2}-{b}^{2}\right)\right]\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{1}{2}\left(a{b}^{2}-{a}^{2}b\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|ab\left(b-a\right)\right|\phantom{\rule{0ex}{0ex}}\ne 0\left(\because a\ne b\ne 0\right)$

Since the area of the triangle formed by the points (*a*, *a*^{2}), (*b*, *b*^{2}) and (0, 0) is not zero, so the given points are not collinear.

#### Page No 6.54:

#### Question 22:

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is $\left(\frac{7}{2},y\right),$ find the value of *y*.

#### Answer:

Let $A\left({x}_{1},{y}_{1}\right)=A\left(2,1\right),B\left({x}_{2},{y}_{2}\right)=B\left(3,-2\right)\mathrm{and}C\left({x}_{3},{y}_{3}\right)=C\left(\frac{7}{2},y\right).$

Now

$\mathrm{Area}\left(\u2206ABC\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}\mathit{-}{y}_{\mathit{2}}\right)\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 5=\frac{1}{2}\left|2\left(-2-y\right)+3\left(y-1\right)+\frac{7}{2}\left(1+2\right)\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 10=\left|-4-2y+3y-3+\frac{21}{2}\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 10=\left|y+\frac{7}{2}\right|\phantom{\rule{0ex}{0ex}}\Rightarrow 10=y+\frac{7}{2}\mathrm{or}-10=y+\frac{7}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{13}{2}\mathrm{or}y=\frac{-27}{2}$

Hence, *y *= $\frac{13}{2}\mathrm{or}\frac{-27}{2}.$

#### Page No 6.54:

#### Question 23:

Prove that the points (a, 0), (0, b) and (1, 1) are collinear if $\frac{1}{a}+\frac{1}{b}=1$.

#### Answer:

The formula for the area ‘*A*’ encompassed by three points, and is given by the formula,

We know area of triangle formed by three points is given by

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(*a,*0)*, B*(0*,b*) and *C*(1*,*1).

$A=\frac{1}{2}\left|a(b-1)+1(0-b)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|ab-a-b\right|$

It is given that

So we have,

Using this in the previously arrived equation for area we have,

Since the area enclosed by the three points is equal to 0, the three points need to be.

#### Page No 6.54:

#### Question 24:

The point A divides the join of *P* (−5, 1) and *Q*(3, 5) in the ratio *k*:1. Find the two values of *k* for which the area of Δ*ABC* where *B* is (1, 5) and *C*(7, −2) is equal to 2 units.

#### Answer:

GIVEN: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio *k*: 1

Coordinates of point B (1, 5) and C (7, −2)

TO FIND: The value of* k*

PROOF: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio *k*: 1

So the coordinates of A are

We know area of triangle formed by three points is given by $\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

Now Area of ΔABC= 2 sq units.

Taking three points A , B (1, 5) and C (7, −2)

Hence

#### Page No 6.54:

#### Question 25:

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x + 3. Find the third vertex.

#### Answer:

GIVEN: The area of triangle is 5.Two of its vertices are (2, 1) and (3, −2). The third vertex lies on *y* = *x*+3

TO FIND: The third vertex.

PROOF: Let the third vertex be (*x*, *y*)

We know area of triangle formed by three points is given by $\u2206=\frac{1}{2}\left|\left({x}_{1}{y}_{2}+{x}_{2}{y}_{3}+{x}_{3}{y}_{1}\right)-\left({x}_{2}{y}_{1}+{x}_{3}{y}_{2}+{x}_{1}{y}_{3}\right)\right|$

Now

Taking three points(*x*, *y*), (2, 1) and (3, −2)

Also it is given the third vertex lies on *y* = *x*+3

Substituting the value in equation (1) and (2) we get

Hence the coordinates of and

#### Page No 6.54:

#### Question 26:

If , prove that the points (*a*, a^{2}), (*b*, *b*^{2}), (*c*, *c*^{2}) can never be collinear.

#### Answer:

GIVEN: If

TO PROVE: that the points can never be collinear.

PROOF:

We know three points are collinear when

Now taking three points

Also it is given that

Hence area of triangle made by these points is never zero. Hence given points are never collinear.

#### Page No 6.54:

#### Question 27:

Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that $\frac{\u2206DBC}{\u2206ABC}=\frac{1}{2},$find *x*.

#### Answer:

GIVEN: four points A (6, 3), B (−3, 5) C (4, −2) and D(*x*, 3*x*) such that

TO FIND: the value of *x*

PROOF:

We know area of the triangles formed by three points is given by

Now

Area of triangle DBC taking D(*x*, 3*x*), B (−3, 5), C (4, −2)

Area of triangle ABC taking, A (6, 3), B (−3, 5), C (4, −2)

Also it is given that

Substituting the values from (1) and (2) we get

#### Page No 6.55:

#### Question 28:

If three points (x_{1}, y_{1}) (x_{2}, y_{2}), (x_{3}, y_{3}) lie on the same line, prove that

$\frac{{y}_{2}-{y}_{3}}{{x}_{2}{x}_{3}}+\frac{{y}_{3}-{y}_{1}}{{x}_{3}{x}_{1}}+\frac{{y}_{1}-{y}_{2}}{{x}_{1}{x}_{2}}=0$

#### Answer:

GIVEN: If three points lie on the same line

TO PROVE:

PROOF:

We know that three points are collinear if

Hence proved.

#### Page No 6.55:

#### Question 29:

Find the area of a parallelogram *ABCD* if three of its vertices are *A*(2, 4), *B*(2 + $\sqrt{3}$, 5) and *C*(2, 6). [CBSE 2013]

#### Answer:

It is given that A(2, 4), B(2 + $\sqrt{3}$, 5) and C(2, 6) are the vertices of the parallelogram ABCD.

We know that the diagonal of a parallelogram divides it into two triangles having equal area.

∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC

Now,

$\mathrm{ar}\left(\u2206\mathrm{ABC}\right)=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|2\left(5-6\right)+\left(2+\sqrt{3}\right)\left(6-4\right)+2\left(4-5\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-2+4+2\sqrt{3}-2\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 2\sqrt{3}\phantom{\rule{0ex}{0ex}}=\sqrt{3}\mathrm{square}\mathrm{units}$

∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC = 2 × $\sqrt{3}$ = 2$\sqrt{3}$ square units

Hence, the area of given parallelogram is 2$\sqrt{3}$ square units.

#### Page No 6.55:

#### Question 30:

Find the value(s) of *k* for which the points (3*k* − 1, *k* − 2), (*k*, *k* − 7) and (*k* − 1, −*k* − 2) are collinear. [CBSE 2014]

#### Answer:

Let A(3*k* − 1, *k* − 2), B(*k*, *k* − 7) and C(*k* − 1, −*k* − 2) be the given points.

The given points are collinear. Then,

$\mathrm{ar}\left(\u2206\mathrm{ABC}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \left(3k-1\right)\left[\left(k-7\right)-\left(-k-2\right)\right]+k\left[\left(-k-2\right)-\left(k-2\right)\right]+\left(k-1\right)\left[\left(k-2\right)-\left(k-7\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(3k-1\right)\left(2k-5\right)+k\left(-2k\right)+5\left(k-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 6{k}^{2}-17k+5-2{k}^{2}+5k-5=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4{k}^{2}-12k=0$

$\Rightarrow 4k\left(k-3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0\mathrm{or}k-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0\mathrm{or}k=3$

Hence, the value of *k* is 0 or 3.

#### Page No 6.55:

#### Question 31:

If the points A(−1, −4), B(*b*, *c*) and C(5, −1) are collinear and 2*b* + *c* = 4, find the values of *b* and *c*. [CBSE 2014]

#### Answer:

The given points A(−1, −4), B(*b*, *c*) and C(5, −1) are collinear.

$\therefore \mathrm{ar}\left(\u2206\mathrm{ABC}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow -1\left[c-\left(-1\right)\right]+b\left[-1-\left(-4\right)\right]+5\left(-4-c\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow -c-1+3b-20-5c=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3b-6c=21\phantom{\rule{0ex}{0ex}}\Rightarrow b-2c=7.....\left(1\right)$

Also, it is given that

2*b* + *c* = 4 .....(2)

Solving (1) and (2), we get

$2\left(7+2c\right)+c=4\phantom{\rule{0ex}{0ex}}\Rightarrow 14+4c+c=4\phantom{\rule{0ex}{0ex}}\Rightarrow 5c=-10\phantom{\rule{0ex}{0ex}}\Rightarrow c=-2$

Putting *c* = −2 in (1), we get

$b-2\times \left(-2\right)=7\phantom{\rule{0ex}{0ex}}\Rightarrow b=7-4=3$

Hence, the respective values of *b* and *c* are 3 and −2.

#### Page No 6.55:

#### Question 32:

If the points A(−2, 1), B(*a*, *b*) and C(4, −1) ae collinear and *a* − *b* = 1, find the values of *a* and *b*. [CBSE 2014]

#### Answer:

The given points A(−2, 1), B(*a*, *b*) and C(4, −1) are collinear.

$\therefore \mathrm{ar}\left(\u2206\mathrm{ABC}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow -2\left[b-\left(-1\right)\right]+a\left(-1-1\right)+4\left(1-b\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2b-2-2a+4-4b=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2a-6b=-2\phantom{\rule{0ex}{0ex}}\Rightarrow a+3b=1.....\left(1\right)$

Also, it is given that

*a* − *b* = 1 .....(2)

Solving (1) and (2), we get

$b+1+3b=1\phantom{\rule{0ex}{0ex}}\Rightarrow 4b=0\phantom{\rule{0ex}{0ex}}\Rightarrow b=0$

Putting *b* = 0 in (1), we get

$a+3\times 0=1\phantom{\rule{0ex}{0ex}}\Rightarrow a=1$

Hence, the respective values of *a* and *b* are 1 and 0.

#### Page No 6.55:

#### Question 33:

If the points $A(1,-2),B(2,3),C(a,2)\mathrm{and}D(-4,-3)$ form a parallelogram , find the value of * a * and height of the parallelogram taking * AB* as base .

#### Answer:

Since diagonals of a parallelogram bisect eachother.

Coordinates of the midpoint of AC = coordinates of the midpoint of BD.

$\Rightarrow \left(\frac{a+1}{2},\frac{2-2}{2}\right)=\left(\frac{-4+2}{2},\frac{-3+3}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{a+1}{2},0\right)=\left(-1,0\right)\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{comparing},\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+1}{2}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow a=-3\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{the}\u2206\mathrm{ABC}\mathrm{is}\phantom{\rule{0ex}{0ex}}A=\frac{1}{2}\left[1\left(3-2\right)+2\left(2+2\right)-3\left(-2-3\right)\right]\phantom{\rule{0ex}{0ex}}A=\frac{1}{2}\left[1+8+15\right]\phantom{\rule{0ex}{0ex}}A=12\mathrm{sq}.\mathrm{units}\phantom{\rule{0ex}{0ex}}$

Since, ABCD is a parallelogram,

Area of ABCD = 2 × area of triangle ABC = 2 × 12 = 24 sq. units

Height of the parallelogram is area of the parallelogram divided by its base.

Base AB is

$AB=\sqrt{{\left(1-2\right)}^{2}+{\left(-2-3\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{1}^{2}+{5}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{26}\phantom{\rule{0ex}{0ex}}\mathrm{Height}=\frac{24}{\sqrt{26}}=\frac{12\sqrt{2}}{\sqrt{26}}$

#### Page No 6.55:

#### Question 34:

$A\left(6,1\right),B(8,2)\mathrm{and}C(9,4)$ are three vertices of a parallelogram *ABCD* . If E is the mid-point of *DC* , find the area of $\u2206$ *ADE*.

#### Answer:

Three vertices are given, then D can be calulated and it comes out to be (7, 3).

Since, E is midpoint of BD.

Therefore, coordinates of E are $\left(\frac{15}{2},\frac{{\displaystyle 5}}{{\displaystyle 2}}\right)$.

Now, vertices of triangle ABE rae (6, 1), (8, 2) and $\left(\frac{15}{2},\frac{{\displaystyle 5}}{{\displaystyle 2}}\right)$.

$\Rightarrow Areaofthe\u2206ABE=\frac{1}{2}\left|\begin{array}{ccc}1& 6& 1\\ 1& 8& 2\\ 1& \frac{15}{2}& \frac{5}{2}\end{array}\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[1\left(20-15\right)-6\left(\frac{5}{2}-2\right)+1\left(\frac{15}{2}-8\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[5-\frac{6}{2}-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{3}{4}aq.units$

#### Page No 6.55:

#### Question 35:

If $D\left(-\frac{1}{5},\frac{5}{2}\right),E(7,3)\mathrm{and}\mathrm{F}\left(\frac{7}{2},\frac{7}{2}\right)$ are the mid-points of sides of $\u2206ABC$, find the area of $\u2206ABC$.

#### Answer:

The midpoint of BC is $D\left(-\frac{1}{2},\frac{5}{2}\right)$,

The midpoint of AB is $F\left(\frac{7}{2},\frac{7}{2}\right)$,

The midpoint of AC is $E\left(7,3\right)$,

Consider the line segment BC,

$\Rightarrow \frac{p+r}{2}=-\frac{1}{2};\frac{q+s}{2}=\frac{5}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow p+r=-1;q+s=5.....\left(i\right)\phantom{\rule{0ex}{0ex}}ConsiderthelinesegmentAB,\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{p+x}{2}=\frac{7}{2};\frac{q+y}{2}=\frac{7}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow p+x=7;q+y=7.....\left(ii\right)$

$ConsiderthelinesegmentAC,\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{r+x}{2}=7;\frac{s+y}{2}=3\phantom{\rule{0ex}{0ex}}\Rightarrow r+x=14;s+y=6.....\left(iii\right)$

Solve (i), (ii) and (iii) to get $A\left(x,y\right)=A\left(11,4\right),B\left(p,q\right)=B\left(-4,3\right),C\left(r,s\right)=C\left(3,2\right)$

Let us assume that BC is base of the triangle,

$\overline{)BC}=\sqrt{{\left(-4-3\right)}^{2}+{\left(3-2\right)}^{2}}=\sqrt{50}\phantom{\rule{0ex}{0ex}}\mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{BC}\mathrm{is}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{x}+4}{-4-3}=\frac{\mathrm{y}-3}{3-2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}+7\mathrm{y}-17=0\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{perpendicular}\mathrm{distance}\mathrm{from}\mathrm{a}\mathrm{point}\mathrm{P}\left({\mathrm{x}}_{1},{\mathrm{y}}_{1}\right)\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{P}=\left|\frac{1\left(11\right)+7\left(4\right)-17}{\sqrt{50}}\right|=\frac{22}{\sqrt{50}}$

The area of the triangle is $A=\frac{1}{2}\times \sqrt{50}\times \frac{22}{\sqrt{50}}=11\mathrm{sq}.\mathrm{units}$

#### Page No 6.61:

#### Question 1:

*Mark the correct alternative in each of the following:*

The distance between the points (cos θ, 0) and (sin θ − cos θ) is

(a) $\sqrt{3}$

(b) $\sqrt{2}$

(c) 2

(d) 1

#### Answer:

We have to find the distance between and.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

So, the answer is (b)

#### Page No 6.61:

#### Question 2:

The distance between the points (*a* cos 25°, 0) and (0, *a* cos 65°) is

(a) *a *

(b) 2*a*

(c) 3*a*

(d) None of these

#### Answer:

We have to find the distance between A(*a* cos 25°, 0) and.

In general, the distance between A and B is given by,

So,

$AB=\sqrt{{\left(0-a\mathrm{cos}25\xb0\right)}^{2}+{\left(a\mathrm{cos}65\xb0-0\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(a\mathrm{cos}25\xb0\right)}^{2}+{\left(a\mathrm{cos}65\xb0\right)}^{2}}$

$\mathrm{cos}25\xb0=\mathrm{sin}65\xb0\mathrm{and}\mathrm{cos}65\xb0=\mathrm{sin}25\xb0$

But according to the trigonometric identity,

Therefore,

So, the answer is (a)

#### Page No 6.61:

#### Question 3:

If x is a positive integer such that the distance between points P (x, 2) and Q (3, −6) is 10 units, then *x* =

(a) 3

(b) −3

(c) 9

(d) −9

#### Answer:

It is given that distance between P (*x*, 2) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (c)

#### Page No 6.61:

#### Question 4:

The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is

(a) a^{2} + b^{2}

(b) a + b

(c) a^{2} − b^{2}

(d) $\sqrt{a2+b2}$

#### Answer:

We have to find the distance between and.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

So, the answer is (d)

#### Page No 6.61:

#### Question 5:

If the distance between the points (4, *p*) and (1, 0) is 5, then *p = *

(a) ± 4

(b) 4

(c) −4

(d) 0* *

#### Answer:

It is given that distance between P (4, *p*) and is 5.

In general, the distance between A and B is given by,

So,

On further simplification,

So,

So the answer is (a)

#### Page No 6.62:

#### Question 6:

A line segment is of length 10 units. If the coordinates of its one end are (2, −3) and the abscissa of the other end is 10, then its ordinate is

(a) 9, 6

(b) 3, −9

(c) −3, 9

(d) 9, −6

#### Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (b)

#### Page No 6.62:

#### Question 7:

The perimeter of the triangle formed by the points (0, 0), (0, 1) and (0, 1) is

(a) 1 ± $\sqrt{2}$

(b) $\sqrt{2}$ + 1

(c) 3

(d) $2+\sqrt{2}$

#### Answer:

We have a triangle whose co-ordinates are A (0, 0); B (1, 0); C (0, 1). So clearly the triangle is right angled triangle, right angled at A. So,

Now apply Pythagoras theorem to get the hypotenuse,

So the perimeter of the triangle is,

Therefore the answer is (d)

#### Page No 6.62:

#### Question 8:

If *A* (2, 2), *B* (−4, −4) and *C* (5, −8) are the vertices of a triangle, than the length of the median through vertex *C* is

(a) $\sqrt{65}$

(b) $\sqrt{117}$

(c) $\sqrt{85}$

(d) $\sqrt{113}$

#### Answer:

We have a triangle in which the co-ordinates of the vertices are A (2, 2) B (−4,−4) and C (5,−8).

In general to find the mid-point of two points and we use section formula as,

Therefore mid-point D of side AB can be written as,

Now equate the individual terms to get,

So co-ordinates of D is (−1,−1)

So the length of median from C to the side AB,

So the answer is (c)

#### Page No 6.62:

#### Question 9:

If three points (0, 0), $\left(3,\sqrt{3}\right)$ and (3, λ) form an equilateral triangle, then λ =

(a) 2

(b) −3

(c) −4

(d) None of these

#### Answer:

We have an equilateral triangle whose co-ordinates are A (0, 0); and.

Since the triangle is equilateral. So,

So,

Cancel out the common terms from both the sides,

Therefore,

So, the answer is (d)

#### Page No 6.62:

#### Question 10:

If the points (*k*, 2*k*), (3*k*, 3*k*) and (3, 1) are collinear, then *k*

(a) $\frac{1}{3}$

(b) $-\frac{1}{3}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

#### Answer:

We have three collinear points.

In general if are collinear then, area of the triangle is 0.

So,

So,

Take out the common terms,

Therefore,

So the answer is (b)

#### Page No 6.62:

#### Question 11:

The coordinates of the point on X-axis which are equidistant from the points (−3, 4) and (2, 5) are

(a) (20, 0)

(b) (−23, 0)

(c) $\left(\frac{4}{5},0\right)$

(d) None of these

#### Answer:

Let the point be A be equidistant from the two given points P (−3, 4) and Q (2, 5).

So applying distance formula, we get,

Therefore,

Hence the co-ordinates of A are

So the answer is- (D) none of these.

#### Page No 6.62:

#### Question 12:

If (−1, 2), (2, −1) and (3, 1) are any three vertices of a parallelogram, then

(a) *a* = 2, *b* = 0

(b)* a*= −2, *b* = 0

(c) *a* = −2, *b *= 6

(d) *a* = 6, *b* = 2

#### Answer:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−1, 2);

B (2,−1) and C(3, 1). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

Now to find the mid-point of two points and we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

So the forth vertex is

#### Page No 6.62:

#### Question 13:

If *A* (5, 3), *B* (11, −5) and *P* (12, *y*) are the vertices of a right triangle right angled at *P*, then *y*=

(a) −2, 4

(b) −2, −4

(c) 2, −4

(d) 2, 4

#### Answer:

Disclaimer: option (b) and (c) are given to be same in the book. So, we are considering option (b) as −2, −4

instead of −2, 4.

We have a right angled triangle whose co-ordinates are A (5, 3); B (11,−5);

. So clearly the triangle is, right angled at A. So,

Now apply Pythagoras theorem to get,

So,

On further simplification we get the quadratic equation as,

Now solve this equation using factorization method to get,

Therefore,

So the answer is (c)

#### Page No 6.62:

#### Question 14:

The area of the triangle formed by (*a*, *b* + *c*), (*b*, *c* + *a*) and (*c*, *a* + *b*)

(a) *a* + *b* + *c*

(b) *abc*

(c) (*a* +* b* + *c*)^{2}

(d) 0

#### Answer:

We have three non-collinear points.

In general if are non-collinear points then are of the triangle formed is given b*y-*

So,

So the answer is (d)

#### Page No 6.62:

#### Question 15:

If (*x* , 2), (−3, −4) and (7, −5) are collinear, then* x* =

(a) 60

(b) 63

(c) −63

(d) −60

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Therefore,

So the answer is (c)

#### Page No 6.62:

#### Question 16:

If points (*t*, 2*t*), (−2, 6) and (3, 1) are collinear, then *t* =

(a) $\frac{3}{4}$

(b) $\frac{4}{3}$

(c) $\frac{5}{3}$

(d) $\frac{3}{5}$

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

So,

Therefore,

So the answer is (b)

#### Page No 6.62:

#### Question 17:

If the area of the triangle formed by the points (*x*, 2*x*), (−2, 6) and (3, 1) is 5 square units , then x =

(a) $\frac{2}{3}$

(b) $\frac{3}{5}$

(c) 3

(d) 5

#### Answer:

We have the co-ordinates of the vertices of the triangle aswhich has an area of 5 sq.units.

In general if are non-collinear points then area of the triangle formed is given b*y-*,

So,

Simplify the modulus function to get,

Therefore,

So the answer is (a)

#### Page No 6.62:

#### Question 18:

If points *(a*, 0), (0, *b*) and (1, 1) are collinear, then $\frac{1}{a}+\frac{1}{b}=$

(a) 1

(b) 2

(c) 0

(d) −1

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Divide both the sides by,

So the answer is (a)

#### Page No 6.62:

#### Question 19:

If the centroid of a triangle is (1, 4) and two of its vertices are (4, −3) and (−9, 7), then the area of the triangle is

(a) 183 sq. units

(b) $\frac{183}{2}$ sq. units

(c) 366 sq. units

(d) $\frac{183}{4}$ sq. units

#### Answer:

The co-ordinates of other two vertices are (4,−3) and (−9, 7)

The co-ordinate of the centroid is (1, 4)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex is (8, 8)

In general if are non-collinear points then are of the triangle formed is given b*y-*,

So,

So the answer is (b)

#### Page No 6.62:

#### Question 20:

The line segment joining points (−3, −4), and (1, −2) is divided by y-axis in the ratio

(a) 1 : 3

(b) 2 : 3

(c) 3 : 1

(d) 2 : 3

#### Answer:

Let P be the point of intersection of *y-*axis with the line segment joining A (−3,−4) and B (1,−2) which divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining and in the ratio *m:n* internally than,

Now we will use section formula as,

Now equate the *x* component on both the sides,

On further simplification,

So *y-*axis divides AB in the ratio

So the answer is (c)

#### Page No 6.62:

#### Question 21:

Find the value of *a* so that the point (3, *a*) lies on the line represented by 2*x* − 3*y** *+ 5 = 0

#### Answer:

If a point is said lie on a line represented by, then the given equation of the line should hold true when the values of the co-ordinates of the points are substituted in it.

Here it is said that the point (3*, a*) lies on the line represented by the equation.

Substituting the co-ordinates of the values in the equation of the line we have,

Thus the value of ‘*a*’ satisfying the given conditions is.

#### Page No 6.63:

#### Question 21:

The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is

(a) −2 : 3

(b) −3 : 2

(c) 3 : 2

(d) 2 : 3

#### Answer:

Here it is said that the point (4, 5)* *divides the points A(2,3) and B(7,8). Substituting these values in the above formula we have,

Equating the individual components we have,

Hence the correct choice is option (d).

#### Page No 6.63:

#### Question 22:

The ratio in which the x-axis divides the segment joining (3, 6) and (12, −3) is

(a) 2: 1

(b) 1 : 2

(c) −2 : 1

(d) 1 : −2

#### Answer:

Let P be the point of intersection of *x-*axis with the line segment joining A (3, 6) and B (12, −3) which divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio m: n internally than,

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So *x-*axis divides AB in the ratio

So the answer is (a)

#### Page No 6.63:

#### Question 23:

If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then *a*^{3}^{ }+ *b*^{3} + *c*^{3} =

(a) *abc*

(b) 0

(c) *a* + *b* + *c*

(d) 3 *abc*

#### Answer:

The co-ordinates of the vertices are (*a, b*); (*b, c*) and (*c, a*)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

Therefore,

We have to find the value of -

Now as we know that if,

Then,

So the answer is (d)

#### Page No 6.63:

#### Question 24:

If Points (1, 2) (−5, 6) and (a, −2) are collinear, then a =

(a) −3

(b) 7

(c) 2

(d) −2

#### Answer:

We have three collinear points.

In general if are collinear then,

So,

So,

Therefore,

So the answer is (b)

#### Page No 6.63:

#### Question 25:

If the centroid of the triangle formed by (7, x) (y, −6) and (9, 10) is at (6, 3), then (x, y) =

(a) (4, 5)

(b) (5, 4)

(c) (−5, −2)

(d) (5, 2)

#### Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is (6, 3)

We know that the co-ordinates of the centroid of a triangle whose vertices are is

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the answer is (d)

#### Page No 6.63:

#### Question 26:

The distance of the point (4, 7) from the x-axis is

(a) 4

(b) 7

(c) 11

(d) $\sqrt{65}$

#### Answer:

The ordinate of a point gives its distance from the *x-*axis.

So, the distance of (4, 7) from *x-*axis is

So the answer is (b)

#### Page No 6.63:

#### Question 27:

The distance of the point (4, 7) from the y-axis is

(a) 4

(b) 7

(c) 11

(d) $\sqrt{65}$

#### Answer:

The distance of a point from *y-*axis is given by abscissa of that point.

So, distance of (4, 7) from *y-*axis is.

So the answer is (a)

#### Page No 6.63:

#### Question 28:

If *P* is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point *Q *on *OY* such that *OP* = *OQ*, are

(a) (0, 3)

(b) (3, 0)

(c) (0, 0)

(d) (0, −3)

#### Answer:

GIVEN: If P is a point on *x* axis such that its distance from the origin is 3 units.

TO FIND: The coordinates of a point Q on OY such that OP= OQ.

On *x* axis y coordinates is 0. Hence the coordinates of point P will be (3, 0) as it is given that the distance from origin is 3 units.

Now then the coordinates of Q on OY such that OP = OQ

On *y* axis *x* coordinates is 0. Hence the coordinates of point Q will be (0, 3)

Hence correct option is (*a*)

#### Page No 6.63:

#### Question 29:

If the points(*x*, 4) lies on a circle whose centre is at the origin and radius is 5, then *x* =

(a) ±5

(b) ±3

(c) 0

(d) ±4

#### Answer:

It is given that the point A(*x*, 4) is at a distance of 5 units from origin O.

So, apply the distance formula to get,

Therefore,

So,

So the answer is (b)

#### Page No 6.63:

#### Question 30:

If the points *P* (*x*, *y*) is equidistant from *A* (5, 1) and *B* (−1*,* 5*)*, then

(a) 5*x* = *y*

(b) *x* = 5*y*

(c) 3*x* = 2*y*

(d) 2*x* = 3*y*

#### Answer:

It is given that is equidistant to the point

So,

So apply distance formula to get the co-ordinates of the unknown value as,

On further simplification we get,

So,

Thus,

So the answer is (c)

#### Page No 6.63:

#### Question 31:

If points *A* (5, *p*) *B* (1, 5), *C* (2, 1) and *D* (6, 2) form a square *ABCD*, then *p* =

(a) 7

(b) 3

(c) 6

(d) 8

#### Answer:

The distance *d* between two points and is given by the formula

In a square all the sides are equal to each other.

Here the four points are A(5,*p*)*, *B(1,5), C(2,1) and D(6,2).

The vertex ‘A’ should be equidistant from ‘B’ as well as *‘*D’

Let us now find out the distances ‘AB’ and ‘AD’.

These two need to be equal.

Equating the above two equations we have,

Squaring on both sides we have,

Hence the correct choice is option (c).

#### Page No 6.63:

#### Question 32:

The coordinates of the circumcentre of the triangle formed by the points *O* (0, 0), *A* (a, 0 and *B* (0, *b*) are

(a) (*a*, *b*)

(b) $\left(\frac{a}{2},\frac{b}{2}\right)$

(c) $\left(\frac{b}{2},\frac{a}{2}\right)$

(d) (*b*, *a*)

#### Answer:

The distance *d* between two points and is given by the formula

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be O(0,0), A(*a*,0) and B(0,*b*).

Let the circumcentre of the triangle be represented by the point R*(x, y)*.

So we have

Equating the first pair of these equations we have,

Squaring on both sides of the equation we have,

Equating another pair of the equations we have,

Squaring on both sides of the equation we have,

Hence the correct choice is option (*b*).

#### Page No 6.63:

#### Question 33:

The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (−3, 4) are

(a) (0, 2)

(b) (3, 0)

(c) (0, 3)

(d) (2, 0)

#### Answer:

TO FIND: The coordinates of a point on *x* axis which lies on perpendicular bisector of line segment joining points (7, 6) and (−3, 4).

Let P(*x*, *y*) be any point on the perpendicular bisector of AB. Then,

PA=PB

On *x*-axis *y* is 0, so substituting *y*=0 we get *x*= 3

Hence the coordinates of point is i.e. option (*b*) is correct

#### Page No 6.63:

#### Question 34:

If the centroid of the triangle formed by the points (3, −5), (−7, 4), (10, −*k*) is at the point (*k* −1), then *k* =

(a) 3

(b) 1

(c) 2

(d) 4

#### Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So the answer is (c)

#### Page No 6.63:

#### Question 35:

If (−2, 1) is the centroid of the triangle having its vertices at (*x* , 0) (5, −2), (−8, *y*), then *x*, *y* satisfy the relation

(a) 3*x* + 8*y* = 0

(b) 3*x* − 8*y* = 0

(c) 8*x* + 3*y* = 0

(d) 8*x* = 3*y*

#### Answer:

We have to find the unknown co-ordinates.

The co-ordinates of vertices are

The co-ordinate of the centroid is (−2, 1)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

It can be observed that (*x*, *y*) = (−3, 5) does not satisfy any of the relations 3*x* + 8*y* = 0, 3*x* − 8*y* = 0, 8*x* + 3*y* = 0 or 8*x* = 3*y*.

#### Page No 6.63:

#### Question 36:

The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are

(a) (3, 0)

(b) (0, 2)

(c) (2, 3)

(d) (3, 2)

#### Answer:

We have to find the co-ordinates of forth vertex of the rectangle ABCD.

We the co-ordinates of the vertices as (0, 0); (2, 0); (0, 3)

Rectangle has opposite pair of sides equal.

When we plot the given co-ordinates of the vertices on a Cartesian plane, we observe that the length and width of the rectangle is 2 and 3 units respectively.

So the co-ordinate of the forth vertex is

So the answer is (c).

#### Page No 6.64:

#### Question 37:

The length of a line segment joining *A* (2, −3) and *B* is 10 units. If the abscissa of *B* is 10 units, then its ordinates can be

(a) 3 or −9

(b) −3 or 9

(c) 6 or 27

(d) −6 or −27

#### Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

So the answer is (a)

#### Page No 6.64:

#### Question 38:

The ratio in which the line segment joining *P* (*x*_{1}, *y*_{1}) and *Q* (*x*_{2},* **y*_{2}) is divided by x-axis is

(a) y_{1} : y_{2}

(b) −y_{1} : y_{2}

(c) x_{1} : x_{2}

(d) −x_{1} : x_{2}

#### Answer:

Let C be the point of intersection of *x-*axis with the line segment joining and which divides the line segment PQ in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio *m:n* internally than,

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So the answer is (b)

#### Page No 6.64:

#### Question 39:

The ratio in which the line segment joining points *A* (*a*_{1}, *b*_{1}) and *B* (*a*_{2}, *b*_{2}) is divided by *y*-axis is

(a) −*a*_{1} : *a*_{2}

(b) *a*_{1}_{ }: *a*_{2}

(c) *b*_{1} : *b*_{2}

(d) −*b*_{1} : *b*_{2}

#### Answer:

Let P be the point of intersection of *y-*axis with the line segment joiningandwhich divides the line segment AB in the ratio.

Now according to the section formula if point a point P divides a line segment joining andin the ratio *m:n* internally than,

Now we will use section formula as,

Now equate the *x* component on both the sides,

On further simplification,

So the answer is (a)

#### Page No 6.64:

#### Question 40:

If the line segment joining the points (3, −4), and (1, 2) is trisected at points *P* (a, −2) and *Q* $\left(\frac{5}{3},b\right)$, Then,

(a) $a=\frac{8}{3},b=\frac{2}{3}$

(b) $a=\frac{7}{3},b=0$

(c) $a=\frac{1}{3},b=1$

(d) $a=\frac{2}{3},b=\frac{1}{3}$

#### Answer:

We have two points A (3,−4) and B (1, 2). There are two points P (*a*,−2) and Q which trisect the line segment joining A and B.

The point P is the point of trisection of the line segment AB. So, P divides AB in the ratio 1: 2

Now we will use section formula to find the co-ordinates of unknown point A as,

Equate the individual terms on both the sides. We get,

Similarly, the point Q is the point of trisection of the line segment AB. So, Q divides AB in the ratio 2: 1

Now we will use section formula to find the co-ordinates of unknown point A as,

Equate the individual terms on both the sides. We get,

So the answer is (b)

#### Page No 6.64:

#### Question 41:

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are

(a) (−6, 7) (b) (6, −7) (c) (6, 7) (d) (−6,−7) [CBSE 2012]

#### Answer:

Let O(−2, 5) be the centre of the given circle and A(2, 3) and B(*x*, *y*) be the end points of a diameter of the circle.

Then, O is the mid-point of AB.

Using mid-point formula, we have

$\therefore \frac{2+x}{2}=-2$ and $\frac{3+y}{2}=5$

$\Rightarrow 2+x=-4$ and $3+y=10$

$\Rightarrow x=-6$ and $y=7$

Thus, the coordinates of the other end of the diameter are (−6, 7).

Hence, the correct answer is option A.

#### Page No 6.64:

#### Question 42:

The coordinates of the point P dividing the line segment joining the points *A* (1, 3) and *B* (4, 6) in the ratio 2 : 1 are

(a) (2, 4) (b) (3, 5) (c) (4, 2) (d) (5, 3) [CBSE 2012]

#### Answer:

It is given that P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1.

Using section formula, we get

Coordinates of P$=\left(\frac{2\times 4+1\times 1}{2+1},\frac{2\times 6+1\times 3}{2+1}\right)=\left(\frac{9}{3},\frac{15}{3}\right)=\left(3,5\right)$

Thus, the coordinates of P are (3, 5).

Hence, the correct answer is option B.

#### Page No 6.64:

#### Question 43:

In Fig. 14.46, the area of Δ*ABC* (in square units) is

(a) 15 (b) 10 (c) 7.5 (d) 2.5 [CBSE 2013]

#### Answer:

The coordinates of A are (1, 3).

∴ Distance of A from the *x*-axis, AD = *y*-coordinate of A = 3 units

The number of units between B and C on the *x*-axis are 5.

∴ BC = 5 units

Now,

Area of ∆ABC = $\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}=\frac{1}{2}\times 5\times 3=\frac{15}{2}=7.5$ square units

Thus, the area of ∆ABC is 7.5 square units.

Hence, the correct answer is option C.

#### Page No 6.64:

#### Question 44:

The point on the *x*-axis which is equidistant from points (−1, 0) and (5, 0) is

(a) (0, 2) (b) (2, 0) (c) (3, 0) (d) (0, 3) [CBSE 2013]

#### Answer:

Let A(−1, 0) and B(5, 0) be the given points. Suppose the required point on the *x*-axis be P(*x*, 0).

It is given that P(*x*, 0) is equidistant from A(−1, 0) and B(5, 0).

∴ PA = PB

⇒ PA^{2 }= PB^{2}

$\Rightarrow {\left[x-\left(-1\right)\right]}^{2}+{\left(0-0\right)}^{2}={\left(x-5\right)}^{2}+{\left(0-0\right)}^{2}$ (Using distance formula)

$\Rightarrow {\left(x+1\right)}^{2}={\left(x-5\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x+1={x}^{2}-10x+25\phantom{\rule{0ex}{0ex}}\Rightarrow 12x=24\phantom{\rule{0ex}{0ex}}\Rightarrow x=2$

Thus, the required point is (2, 0).

Hence, the correct answer is option B.

#### Page No 6.65:

#### Question 45:

If A(4, 9), B(2, 3) and C(6, 5) are the vertices of ∆ABC, then the length of median through C is

(a) 5 units (b) $\sqrt{10}$ units (c) 25 units (d) 10 units [CBSE 2014]

#### Answer:

It is given that A(4, 9), B(2, 3) and C(6, 5) are the vertices of ∆ABC.

Let CD be the median of ∆ABC through C. Then, D is the mid-point of AB.

Using mid-point formula, we get

Coordinates of D = $\left(\frac{4+2}{2},\frac{9+3}{2}\right)=\left(\frac{6}{2},\frac{12}{2}\right)=\left(3,6\right)$

∴ Length of the median, AD

$=\sqrt{{\left(6-3\right)}^{2}+{\left(5-6\right)}^{2}}\left(\mathrm{Using}\mathrm{distance}\mathrm{formula}\right)\phantom{\rule{0ex}{0ex}}=\sqrt{{3}^{2}+{\left(-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{10}\mathrm{units}$

Thus, the length of the required median is $\sqrt{10}$ units.

Hence, the correct answer is option B.

#### Page No 6.65:

#### Question 46:

If P(2, 4), Q(0, 3), R(3, 6) and S(5, *y*) are the vertices of a parallelogram PQRS, then the value of *y* is

(a) 7 (b) 5 (c) −7 (d) −8 [CBSE 2014]

#### Answer:

It is given that P(2, 4), Q(0, 3), R(3, 6) and S(5, *y*) are the vertices of a parallelogram PQRS.

Join PR and QS, intersecting each other at O.

We know that the diagonals of the parallelogram bisect each other. So, O is the mid-point of PR and QS.

Coordinates of mid-point of PR = $\left(\frac{2+3}{2},\frac{4+6}{2}\right)=\left(\frac{5}{2},\frac{10}{2}\right)=\left(\frac{5}{2},5\right)$

Coordinates of mid-point of QS = $\left(\frac{0+5}{2},\frac{3+y}{2}\right)=\left(\frac{5}{2},\frac{3+y}{2}\right)$

Now, these points coincides at the point O.

$\therefore \left(\frac{5}{2},\frac{3+y}{2}\right)=\left(\frac{5}{2},5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3+y}{2}=5\phantom{\rule{0ex}{0ex}}\Rightarrow 3+y=10\phantom{\rule{0ex}{0ex}}\Rightarrow y=7$

Thus, the value of *y* is 7.

Hence, the correct answer is option A.

#### Page No 6.65:

#### Question 47:

If A(*x*, 2), B(−3, −4) and C(7, −5) are collinear, then the value of *x* is

(a) −63 (b) 63 (c) 60 (b) −60 [CBSE 2014]

#### Answer:

The given points A(*x*, 2), B(−3, −4) and C(7, −5) are collinear.

$\therefore \mathrm{ar}\left(\u2206\mathrm{ABC}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)=0$

$\Rightarrow x\left[-4-\left(-5\right)\right]+\left(-3\right)\left(-5-2\right)+7\left[2-\left(-4\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow x+21+42=0\phantom{\rule{0ex}{0ex}}\Rightarrow x+63=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-63$

Thus, the value of *x* is −63.

Hence, the correct answer is option A.

#### Page No 6.65:

#### Question 48:

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(a) 7 + $\sqrt{5}$ (b) 5 (c) 10 (d) 12 [CBSE 2014]

#### Answer:

Let A(0, 4), O(0, 0) and B(3, 0) be the vertices of ∆AOB.

Using distance formula, we get

OA = $\sqrt{{\left(0-0\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{16}=4$ units

OB = $\sqrt{{\left(3-0\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{9}=3$ units

AB = $\sqrt{{\left(3-0\right)}^{2}+{\left(0-4\right)}^{2}}=\sqrt{9+16}=\sqrt{25}=5$ units

∴ Perimeter of ∆AOB = OA + OB + AB = 4 + 3 + 5 = 12 units

Thus, the required perimeter of the triangle is 12 units.

Hence, the correct answer is option D.

#### Page No 6.65:

#### Question 49:

If the point *P *(2, 1 ) lies on the line segment joining points *A *(4,20 and *B *(8, 4) , then

(a) $AP=\frac{1}{3}AB$ (b) *AP* = *BP * (C) *PB* = $\frac{1}{3}AB$ (D) $AP=\frac{1}{2}AB$

#### Answer:

Use section formula for finding out the ratio in which P divided the line segment AB.

$\left(2,1\right)=\left(\frac{{m}_{1}\left(8\right)+{m}_{2}\left(4\right)}{{m}_{1}+{m}_{2}},\frac{{m}_{1}\left(4\right)+{m}_{2}\left(2\right)}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2=\frac{8{m}_{1}+4{m}_{2}}{{m}_{1}+{m}_{2}};1=\frac{4{m}_{1}+2{m}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{m}_{1}+{m}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{m}_{1}}{{m}_{2}}=-\frac{1}{3}\phantom{\rule{0ex}{0ex}}\therefore thepointPdividedABinration1:3externally.\phantom{\rule{0ex}{0ex}}AP:PB=1:3\phantom{\rule{0ex}{0ex}}\Rightarrow AP:AB=1:2\phantom{\rule{0ex}{0ex}}\Rightarrow AP=\frac{1}{2}AB$

#### Page No 6.65:

#### Question 50:

A line intersects the* y*-axis and *x*-axis at *P* and* Q *, respectively. If (2, $-$5) is the mid-point of* PQ*, then the coordinates of* P* and *Q* are, respectively

(a) (0, $-$5) and (2, 0) (b) (0, 10) and ($-$4, 0)

(c) (0, 4) and ($-$10, 0 ) (d) (0, $-$10) and (4 , 0)

#### Answer:

A line intersects the *y* axis, then the coordinates of P are (0, *y*) and *x *axis then the coordinates are Q(*x*, 0).

Therefore by section formula,

$\left(\frac{x+0}{2},\frac{0+y}{2}\right)=\left(2,-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{x}{2},\frac{y}{2}\right)=\left(2,-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{2}=2,\frac{y}{2}=-5\phantom{\rule{0ex}{0ex}}\Rightarrow x=4,y=-10$

Hence the coordinates of P are (0, −10) and that of Q are (4, 0).

#### Page No 6.65:

#### Question 1:

The distance of the point (2, 3) from *x-*axis is ________.

#### Answer:

The distance between the point (2, 3) and *x-*axis can be determined by assuming a point (2, 0) on *x-*axis.

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between (2, 3) and (2, 0)$=\sqrt{{\left(2-2\right)}^{2}+{\left(0-3\right)}^{2}}$

= $\sqrt{{\left(3\right)}^{2}}$

= $\sqrt{9}$

= 3

Hence, the distance of the point (2, 3) from *x-*axis is __3 units__.

#### Page No 6.65:

#### Question 2:

The distance of the point (–4, 7) from *y-*axis is _________.

#### Answer:

The distance between the point (–4, 7) and *y-*axis can be determined by assuming a point (0, 7) on *y-*axis.

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between (–4, 7) and (0, 7)$=\sqrt{{\left(0-\left(-4\right)\right)}^{2}+{\left(7-7\right)}^{2}}$

= $\sqrt{{\left(-4\right)}^{2}}$

= $\sqrt{16}$

= 4

Hence, the distance of the point (–4, 7) from *y-*axis is __4 units__.

#### Page No 6.65:

#### Question 3:

The image of the point (3, –5) in the *x-*axis has the coordinates __________.

#### Answer:

The image of (*x*, *y*) in the *x-*axis is (*x*, –*y*).

Thus, the image of the point (3, –5) in the *x-*axis is (3, 5).

Hence, the image of the point (3, –5) in the *x-*axis has the coordinates __(3, 5)__.

#### Page No 6.65:

#### Question 4:

The coordinates of the image of the point (–4, 5) in *y-*axis are _________.

#### Answer:

The image of (*x*, *y*) in the *y-*axis is (–*x*, *y*).

Thus, the image of the point (–4, 5) in *y-*axis is (4, 5).

Hence, the coordinates of the image of the point (–4, 5) in *y-*axis are __(4, 5)__.

#### Page No 6.65:

#### Question 5:

The distance between the points (0, 5) and (5, 0) is __________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between the points (0, 5) and (5, 0)$=\sqrt{{\left(5-0\right)}^{2}+{\left(0-5\right)}^{2}}$

$=\sqrt{{\left(5\right)}^{2}+{\left(-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25+25}\phantom{\rule{0ex}{0ex}}=\sqrt{50}\phantom{\rule{0ex}{0ex}}=5\sqrt{2}$

Hence, the distance between the points (0, 5) and (5, 0) is $\overline{)5\sqrt{2}\mathrm{units}}.$

#### Page No 6.65:

#### Question 6:

If the distance between the points (2, –2) and (–1, *x*) is 5, then the values of *x* are __________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between the points (2, –2) and (–1, *x*)$=\sqrt{{\left(-1-2\right)}^{2}+{\left(x-\left(-2\right)\right)}^{2}}$

$\Rightarrow 5=\sqrt{{\left(-3\right)}^{2}+{\left(x+2\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(5\right)}^{2}={\left(\sqrt{9+{\left(x+2\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25=9+{\left(x+2\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25=9+{x}^{2}+4+4x\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+4x+13-25=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+4x-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+6x-2x-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x+6\right)-2\left(x+6\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-2\right)\left(x+6\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=2,-6$

Hence, the values of *x* are __2 and –6__.

#### Page No 6.65:

#### Question 7:

The area of the triangle with vertices at *A*(3, 0), *B*(7, 0) and *C*(8, 4) is ___________.

#### Answer:

Area of the triangle formed by the vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}$

$\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$

Now, the given vertices are *A*(3, 0), *B*(7, 0) and *C*(8, 4)

Therefore,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|3\left(0-4\right)+7\left(4-0\right)+8\left(0-0\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|3\left(-4\right)+7\left(4\right)+8\left(0\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-12+28+0\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|16\right|\phantom{\rule{0ex}{0ex}}=8$

Hence, the area of a triangle is __8 square units__.

#### Page No 6.65:

#### Question 8:

The quadrant in which the point dividing the line segment joining the points (7, –6) and (3, 4) internally in the ratio 1 : 2 is _________.

#### Answer:

Let *P*(*x*, *y*) be the point dividing the line segment joining the points (7, –6) and (3, 4) internally in the ratio 1 : 2

Section formula: if the point (*x*, *y*) divides the line segment joining the points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) internally in the ratio *m *:* n*, then the coordinates (*x*, *y*) = $\left(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$

Therefore, using section formula, the coordinates of *P* are:

$\left(x,y\right)=\left(\frac{1\left(3\right)+2\left(7\right)}{1+2},\frac{1\left(4\right)+2\left(-6\right)}{1+2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{3+14}{3},\frac{4-12}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{3},\frac{-8}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{17}{3},-\frac{8}{3}\right)$

Hence, the quadrant in which the point dividing the line segment joining the points (7, –6) and (3, 4) internally in the ratio 1 : 2 is __IV quadrant__.

#### Page No 6.66:

#### Question 9:

The *x*-coordinate of the point lying on the perpendicular bisector of the line segment joining the points* A*(–2, –5) and *B*(2, 5) is_______.

#### Answer:

Perpendicular bisectors divides the line segment in two equal parts.

$\mathrm{Mid}\mathrm{point}\mathrm{of}\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right).$

Therefore, the mid point of *A*(–2, –5) and *B*(2, 5)

$=\left(\frac{-2+2}{2},\frac{-5+5}{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{0}{2},\frac{0}{2}\right)\phantom{\rule{0ex}{0ex}}=\left(0,0\right)$

Hence, the *x*-coordinate of the point lying on the perpendicular bisector of the line segment joining the points* A*(–2, –5) and *B*(2, 5) is __0__.

#### Page No 6.66:

#### Question 10:

If* P*(*a*/3,4) is the mid point of the line segment joining the points *Q*(–6, 5) and *R*(–2, 3), then the value of '*a*' is _________.

#### Answer:

$\mathrm{Mid}\mathrm{point}\mathrm{of}\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right).$

*P*(*a*/3,4) is the mid point of the line segment joining the points *Q*(–6, 5) and *R*(–2, 3).

Therefore, using mid point formula the coordinates of *P* are:

$\left(\frac{a}{3},4\right)=\left(\frac{-6-2}{2},\frac{5+3}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{a}{3},4\right)=\left(\frac{-8}{2},\frac{8}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{a}{3},4\right)=\left(-4,4\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{3}=-4\phantom{\rule{0ex}{0ex}}\Rightarrow a=-12$

Hence, the value of '*a*' is __–12__.

#### Page No 6.66:

#### Question 11:

The coordinates of the point which is equidistant from the vertices of the triangle formed by the points *O*(0, 0), *A*(*a*, 0) and *B*(0, *b*), are _________.

#### Answer:

Circumcentre is the point which is equidistant from the vertices of the triangle.

Let the point be *C*(*x*, *y*).

Thus, *OC* = *AC* = *BC*

$\Rightarrow {\left(OC\right)}^{2}={\left(AC\right)}^{2}={\left(BC\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},{\left(OC\right)}^{2}={\left(AC\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\sqrt{{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(x-a\right)}^{2}+{\left(y-0\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(x-a\right)}^{2}+{\left(y-0\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x\right)}^{2}+{\left(y\right)}^{2}={\left(x-a\right)}^{2}+{\left(y\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}={x}^{2}+{a}^{2}-2ax\phantom{\rule{0ex}{0ex}}\Rightarrow 2ax={a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=a\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{a}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},{\left(OC\right)}^{2}={\left(BC\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\sqrt{{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(x-0\right)}^{2}+{\left(y-b\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(x-0\right)}^{2}+{\left(y-b\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x\right)}^{2}+{\left(y\right)}^{2}={\left(x\right)}^{2}+{\left(y-b\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}^{2}={y}^{2}+{b}^{2}-2by\phantom{\rule{0ex}{0ex}}\Rightarrow 2by={b}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2y=b\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{b}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x=\frac{a}{2}\mathrm{and}y=\frac{b}{2}.$

Hence, the coordinates of the point which is equidistant from the vertices of the triangle formed by the points *O*(0, 0), *A*(*a*, 0) and *B*(0, *b*), are $\overline{)\left(\frac{a}{2},\frac{b}{2}\right)}.$

#### Page No 6.66:

#### Question 12:

If the points (*a*, 0), (0, *b*) and (1, 1) are collinear, then $\frac{1}{a}+\frac{1}{b}$ = ___________.

#### Answer:

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}$ $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given points (*a*, 0), (0, *b*) and (1, 1) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|a\left(b-1\right)+\left(0\right)\left(1-0\right)+1\left(0-b\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|ab-a+0-b\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left|ab-a-b\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow ab-a-b=0\phantom{\rule{0ex}{0ex}}\Rightarrow ab=a+b\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=ab\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}ab,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{ab}+\frac{b}{ab}=\frac{ab}{ab}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{b}+\frac{1}{a}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, $\frac{1}{a}+\frac{1}{b}$ = __1__.

#### Page No 6.66:

#### Question 13:

If the distance of the point (4, *a*) from *x*-axis is half its distance from *y*-axis, then ___________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

The distance between the point (4, *a*) and *x-*axis can be determined by assuming a point (4, 0) on *x-*axis.

Therefore, the distance between (4, *a*) and (4, 0)$=\sqrt{{\left(4-4\right)}^{2}+{\left(0-a\right)}^{2}}$

$=\sqrt{{\left(a\right)}^{2}}\phantom{\rule{0ex}{0ex}}=a$

Also, the distance between the point (4, *a*) and *y-*axis can be determined by assuming a point (0, *a*) on *y-*axis.

Therefore, the distance between (4, *a*) and (0, *a*)$=\sqrt{{\left(4-0\right)}^{2}+{\left(a-a\right)}^{2}}$

$=\sqrt{{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=4$

According to the question,

Distance of the point (4, *a*) from *x*-axis is half its distance from *y*-axis.

$\Rightarrow a=\frac{1}{2}\left(4\right)\phantom{\rule{0ex}{0ex}}\Rightarrow a=2$

Hence, if the distance of the point (4, *a*) from *x*-axis is half its distance from *y*-axis, then __the value of a is 2__.

#### Page No 6.66:

#### Question 14:

If the distance between the points *A*(–3, –14) and *B*(*a,* –5) is 9 units, then *a *= __________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between the points *A*(–3, –14) and *B*(*a,* –5)$=\sqrt{{\left(a-\left(-3\right)\right)}^{2}+{\left(-5-\left(-14\right)\right)}^{2}}$

$\Rightarrow 9=\sqrt{{\left(a+3\right)}^{2}+{\left(-5+14\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(9\right)}^{2}={\left(\sqrt{{\left(a+3\right)}^{2}+{\left(9\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 81={\left(a+3\right)}^{2}+81\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(a+3\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=-3$

Hence, *a *= __–3__.

#### Page No 6.66:

#### Question 15:

If the points (5, 1), (–2, –3) and (8, 2*m*) are collinear, then *m* = ___________.

#### Answer:

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}$ $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$.

Now, the given points (5, 1), (–2, –3) and (8, 2*m*) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|5\left(-3-2m\right)+\left(-2\right)\left(2m-1\right)+8\left(1-\left(-3\right)\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left|-15-10m-4m+2+8\left(1+3\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left|-13-14m+8\left(4\right)\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left|-13-14m+32\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left|-14m+19\right|=0\phantom{\rule{0ex}{0ex}}\Rightarrow 19-14m=0\phantom{\rule{0ex}{0ex}}\Rightarrow 14m=19\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{19}{14}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, *m* = $\overline{)\frac{19}{14}}.$

#### Page No 6.66:

#### Question 16:

The ratio in which the point $P\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points *A*(1/2, 3/2) and* B*(2, –5) is _________.

#### Answer:

Section formula: if the point (*x*, *y*) divides the line segment joining the points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) internally in the ratio *k *:* *1, then the coordinates (*x*, *y*) = $\left(\frac{k{x}_{2}+{x}_{1}}{k+1},\frac{k{y}_{2}+{y}_{1}}{k+1}\right)$

Let the point $P\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points *A*(1/2, 3/2) and* B*(2, –5) in the ratio *k *:* *1.

Therefore, using section formula, the coordinates of *P* are:

$\left(\frac{3}{4},\frac{5}{12}\right)=\left(\frac{k\left(2\right)+1\left({\displaystyle \frac{1}{2}}\right)}{k+1},\frac{k\left(-5\right)+1\left({\displaystyle \frac{3}{2}}\right)}{k+1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{3}{4},\frac{5}{12}\right)=\left(\frac{2k+{\displaystyle \frac{1}{2}}}{k+1},\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{3}{4},\frac{5}{12}\right)=\left(\frac{{\displaystyle \frac{4k+1}{2}}}{k+1},\frac{{\displaystyle \frac{-10k+3}{2}}}{k+1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{3}{4},\frac{5}{12}\right)=\left(\frac{{\displaystyle 4k+1}}{2k+2},\frac{{\displaystyle -10k+3}}{2k+2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle 4k+1}}{2k+2}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\left(4k+1\right)=3\left(2k+2\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 16k+4=6k+6\phantom{\rule{0ex}{0ex}}\Rightarrow 10k=2\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{2}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{1}{5}$

Hence, the ratio in which the point $P\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points *A*(1/2, 3/2) and* B*(2, –5) is __1 : 5__.

#### Page No 6.66:

#### Question 17:

The *x*-axis divides the line segment joining the points (–4, –6) and (–1, 7) in the ratio _________.

#### Answer:

Section formula: if the point (*x*, *y*) divides the line segment joining the points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) internally in the ratio *k *:* *1, then the coordinates (*x*, *y*) = $\left(\frac{k{x}_{2}+{x}_{1}}{k+1},\frac{k{y}_{2}+{y}_{1}}{k+1}\right)$

Let the point *P*(*x*, 0) divides the line segment joining the points (–4, –6) and (–1, 7) in the ratio *k *:* *1.

Therefore, using section formula, the coordinates of *P* are:

$\left(x,0\right)=\left(\frac{k\left(-1\right)+1\left({\displaystyle -4}\right)}{k+1},\frac{k\left(7\right)+1\left({\displaystyle -6}\right)}{k+1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x,0\right)=\left(\frac{-k{\displaystyle -}{\displaystyle 4}}{k+1},\frac{7k-6}{k+1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{-k{\displaystyle -}{\displaystyle 4}}{k+1}\mathrm{and}0=\frac{7k-6}{k+1}\phantom{\rule{0ex}{0ex}}\Rightarrow 0=\frac{7k-6}{k+1}\phantom{\rule{0ex}{0ex}}\Rightarrow 7k-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 7k=6\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{6}{7}$

Hence, the *x*-axis divides the line segment joining the points (–4, –6) and (–1, 7) in the ratio __6 : 7__.

#### Page No 6.66:

#### Question 18:

The values of *y* for which the point (2, –4) is equidistant from the points (3, 8) and (–10, *y*), are _________.

#### Answer:

Let the point *P*(2, –4) is equidistant from the points *A*(3, 8) and *B*(–10, *y*).

Thus, *AP* = *BP*

$\Rightarrow {\left(AP\right)}^{2}={\left(BP\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\sqrt{{\left(2-3\right)}^{2}+{\left(-4-8\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(2-\left(-10\right)\right)}^{2}+{\left(-4-y\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(-1\right)}^{2}+{\left(-12\right)}^{2}={\left(2+10\right)}^{2}+{\left(-4-y\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1+144={\left(12\right)}^{2}+{\left(4+y\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1+144=144+16+{y}^{2}+8y\phantom{\rule{0ex}{0ex}}\Rightarrow 1=16+{y}^{2}+8y\phantom{\rule{0ex}{0ex}}\Rightarrow {y}^{2}+8y+15=0\phantom{\rule{0ex}{0ex}}\Rightarrow {y}^{2}+5y+3y+15=0\phantom{\rule{0ex}{0ex}}\Rightarrow y\left(y+5\right)+3\left(y+5\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(y+3\right)\left(y+5\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow y=-5,-3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},y=-5,-3.$

Hence, the values of *y* for which the point (2, –4) is equidistant from the points (3, 8) and (–10, *y*), are __–5 and –3__.

#### Page No 6.66:

#### Question 19:

If the distance between the points (4, *p*) and (1, 0) is 5, then the values of *p* are __________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between the points (4, *p*) and (1, 0)$=\sqrt{{\left(1-4\right)}^{2}+{\left(0-p\right)}^{2}}$

$\Rightarrow 5=\sqrt{{\left(-3\right)}^{2}+{\left(-p\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(5\right)}^{2}={\left(\sqrt{9+{p}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25=9+{p}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}=25-9\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}=16\phantom{\rule{0ex}{0ex}}\Rightarrow p=\pm 4$

Hence, the values of *p* are __4 and −4__.

#### Page No 6.66:

#### Question 20:

The coordinates of the point equidistant from the vertices *O*(0, 0), *A*(6, 0) and *B*(0, 8) of Δ*OAB* are ___________.

#### Answer:

Circumcentre is the point which is equidistant from the vertices of the triangle.

Let the point be *C*(*x*, *y*).

Thus, *OC* = *AC* = *BC*

$\Rightarrow {\left(OC\right)}^{2}={\left(AC\right)}^{2}={\left(BC\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},{\left(OC\right)}^{2}={\left(AC\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\sqrt{{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(x-6\right)}^{2}+{\left(y-0\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(x-6\right)}^{2}+{\left(y-0\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x\right)}^{2}+{\left(y\right)}^{2}={\left(x-6\right)}^{2}+{\left(y\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}={x}^{2}+36-12x\phantom{\rule{0ex}{0ex}}\Rightarrow 12x=36\phantom{\rule{0ex}{0ex}}\Rightarrow x=3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},{\left(OC\right)}^{2}={\left(BC\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\sqrt{{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(x-0\right)}^{2}+{\left(y-8\right)}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(x-0\right)}^{2}+{\left(y-8\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x\right)}^{2}+{\left(y\right)}^{2}={\left(x\right)}^{2}+{\left(y-8\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}^{2}={y}^{2}+64-16y\phantom{\rule{0ex}{0ex}}\Rightarrow 16y=64\phantom{\rule{0ex}{0ex}}\Rightarrow y=4\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x=3\mathrm{and}y=4.$

Hence, coordinates of the point equidistant from the vertices *O*(0, 0), *A*(6, 0) and *B*(0, 8) of Δ*OAB* are __(3, 4)__.

#### Page No 6.66:

#### Question 21:

The number of points on *x*-axis which are at a distance of $2\sqrt{5}$ from the point (7, –4), is __________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Let the point on *x*-axis be *P*(*x*, 0) which is at a distance of $2\sqrt{5}$ from the point (7, –4).

Therefore, the distance between the points (*x*, 0) and (7, –4)$=\sqrt{{\left(7-x\right)}^{2}+{\left(-4-0\right)}^{2}}$

$\Rightarrow 2\sqrt{5}=\sqrt{{\left(7-x\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(2\sqrt{5}\right)}^{2}={\left(\sqrt{{\left(7-x\right)}^{2}+16}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 20={\left(7-x\right)}^{2}+16\phantom{\rule{0ex}{0ex}}\Rightarrow 20=49+{x}^{2}-14x+16\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-14x+65-20=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-14x+45=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-9x-5x+45=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-9\right)-5\left(x-9\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-5\right)\left(x-9\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=5,9$

Hence, the number of points on *x*-axis which are at a distance of $2\sqrt{5}$ from the point (7, –4), is __2__.

#### Page No 6.66:

#### Question 22:

The distance between the points (1, 0) and (2, cot θ) is _________.

#### Answer:

The distance between $\left({x}_{1},{y}_{1}\right)\mathrm{and}\left({x}_{2},{y}_{2}\right)\mathrm{is}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$.

Therefore, the distance between the points (1, 0) and (2, cot*θ*)$=\sqrt{{\left(2-1\right)}^{2}+{\left(\mathrm{cot}\theta -0\right)}^{2}}$

$=\sqrt{{\left(1\right)}^{2}+{\left(\mathrm{cot}\theta \right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+{\left(\mathrm{cot}\theta \right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\mathrm{Cosec}\theta \right)}^{2}}\phantom{\rule{0ex}{0ex}}=\mathrm{Cosec}\theta $

Hence, the distance between the points (1, 0) and (2, cot*θ*) is __cosec θ__.

#### Page No 6.66:

#### Question 23:

If the centroid of the triangle whose vertices are (2, 4), (3,* a*), (4, 2) is (*a*, 3), then *a =* _________.

#### Answer:

$\mathrm{Centroid}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{vertices}\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right).\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Centroid}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{vertices}\left(2,4\right),\left(3,a\right)\mathrm{and}\left(4,2\right)=\left(a,3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{2+3+4}{3},\frac{4+a+2}{3}\right)=\left(a,3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{9}{3},\frac{6+a}{3}\right)=\left(a,3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(3,\frac{6+a}{3}\right)=\left(a,3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 3=a$

Hence, *a =* __3__.

#### Page No 6.66:

#### Question 24:

It (2, –2) and (5, 2) are opposite vertices of a square, then the length of the side of the square is ___________.

#### Answer:

Length of the opposite vertices of a square = Length of the diagonal

$\therefore \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{diagonal}=\sqrt{{\left(5-2\right)}^{2}+{\left(2-\left(-2\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(3\right)}^{2}+{\left(2+2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(3\right)}^{2}+{\left(4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+16}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5$

But, Length of the diagonal = $\sqrt{2}$ (Length of the side of the square)

$\therefore 5=\sqrt{2}\times \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{side}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{side}=\frac{5}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{5}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{5\sqrt{2}}{2}$

Hence, the length of the side of the square is $\overline{)\frac{5\sqrt{2}}{2}\mathrm{units}}.$

#### Page No 6.66:

#### Question 25:

If the centroid of the triangle formed by the points (*a, b*), (1, *a*) and (*b*, 1) is at the origin, then $\frac{{a}^{3}+{b}^{3}+1}{ab}=$ ____________.

#### Answer:

$\mathrm{Centroid}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{vertices}\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\mathrm{and}\left({x}_{3},{y}_{3}\right)\mathrm{is}\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right).\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Centroid}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{vertices}\left(a,b\right),\left(1,a\right)\mathrm{and}\left(b,1\right)=\left(0,0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{a+1+b}{3},\frac{b+a+1}{3}\right)=\left(0,0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+b+1}{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow a+b+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow a+b=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\frac{{a}^{3}+{b}^{3}+1}{ab}=\frac{{\left(a+b\right)}^{3}-3ab\left(a+b\right)+1}{ab}\phantom{\rule{0ex}{0ex}}=\frac{{\left(-1\right)}^{3}-3ab\left(-1\right)+1}{ab}\left(\because a+b=-1\right)\phantom{\rule{0ex}{0ex}}=\frac{-1+3ab+1}{ab}\phantom{\rule{0ex}{0ex}}=\frac{3ab}{ab}\phantom{\rule{0ex}{0ex}}=3$

Hence, $\frac{{a}^{3}+{b}^{3}+1}{ab}=$__3__.

#### Page No 6.67:

#### Question 1:

Write the distance between the points A (10 cos θ, 0) and B (0, 10 sin θ).

#### Answer:

We have to find the distance between A and B.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

Therefore,

#### Page No 6.67:

#### Question 2:

Write the perimeter of the triangle formed by the points *O* (0, 0), *A* (*a*, 0) and B (0, b).

#### Answer:

The distance *d* between two points and is given by the formula

The perimeter of a triangle is the sum of lengths of its sides.

The three vertices of the given triangle are O(0*, *0), A(*a, *0) and B(0*, b*).

Let us now find the lengths of the sides of the triangle.

The perimeter ‘*P*’ of the triangle is thus,

Thus the perimeter of the triangle with the given vertices is.

#### Page No 6.67:

#### Question 3:

Write the ratio in which the line segment joining points (2, 3) and (3, −2) is divided by X axis.

#### Answer:

Let P be the point of intersection of *x-*axis with the line segment joining A (2, 3) and B (3,−2) which divides the line segment AB in the ratio.

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So *x-*axis divides AB in the ratio

#### Page No 6.67:

#### Question 4:

What is the distance between the points (5 sin 60°, 0) and (0, 5 sin 30°)?

#### Answer:

We have to find the distance between A and B.

In general, the distance between A and B is given by,

So,

But according to the trigonometric identity,

And,

Therefore,

#### Page No 6.67:

#### Question 5:

If *A* (−1, 3) , *B*(1, −1) and *C* (5, 1) are the vertices of a triangle *ABC*, what is the length of the median through vertex *A*?

#### Answer:

We have a triangle in which the co-ordinates of the vertices are A (−1, 3) B (1,−1) and

C (5, 1). In general to find the mid-point of two pointsand we use section formula as,

Therefore mid-point D of side BC can be written as,

Now equate the individual terms to get,

So co-ordinates of D is (3, 0)

So the length of median from A to the side BC,

#### Page No 6.67:

#### Question 6:

If the distance between points (x, 0) and (0, 3) is 5, what are the values of x?

#### Answer:

We have to find the unknown *x* using the distance between A and B which is 5.In general, the distance between A and B is given by,

So,

Squaring both the sides we get,

So,

#### Page No 6.67:

#### Question 7:

What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4)?

#### Answer:

The given triangle is a right angled triangle, right angled at O. the co-ordinates of the vertices are O (0, 0) A (6, 0) and B (0, 4).

So,

Altitude is 6 units and base is 4 units.

Therefore,

#### Page No 6.67:

#### Question 8:

Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.

#### Answer:

Let P be the point which divide the line segment joining A (2, 3) and B (3, 4) in the ratio 1: 5.

Now we will use section formula as,

So co-ordinate of P is

#### Page No 6.67:

#### Question 9:

If the centroid of the triangle formed by points P (a, b), Q(b, c) and R (c, a) is at the origin, what is the value of a + b + c?

#### Answer:

The co-ordinates of the vertices are (*a, b*); (*b, c*) and (*c, a*)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

Therefore,

#### Page No 6.67:

#### Question 10:

In Q. No. 9, what is the value of $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}$?

#### Answer:

The co-ordinates of the vertices are (*a, b*); (*b, c*) and (*c, a*)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

Therefore,

We have to find the value of -

Multiply and divide it by to get,

Now as we know that if,

Then,

So,

#### Page No 6.67:

#### Question 11:

Write the coordinates of a point on X-axis which is equidistant from the points (−3, 4) and (2, 5).

#### Answer:

The distance *d* between two points and is given by the formula

Here we are to find out a point on the *x*−axis which is equidistant from both the points

A(*-*3*,*4) and B(2*,*5).

Let this point be denoted as C(*x, y*).

Since the point lies on the *x*-axis the value of its ordinate will be 0. Or in other words we have.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

We know that both these distances are the same. So equating both these we get,

Squaring on both sides we have,

Hence the point on the *x*-axis which lies at equal distances from the mentioned points is.

#### Page No 6.67:

#### Question 12:

If the mid-point of the segment joining *A* (*x*, *y* + 1) and *B* (*x* + 1, *y* + 2) is *C* $\left(\frac{3}{2},\frac{5}{2}\right)$, find *x*, *y.*

#### Answer:

It is given that mid-point of line segment joining A and B is C

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the components separately to get,

So,

Similarly,

So,

#### Page No 6.67:

#### Question 13:

Two vertices of a triangle have coordinates (−8, 7) and (9, 4) . If the centroid of the triangle is at the origin, what are the coordinates of the third vertex?

#### Answer:

The co-ordinates of other two vertices are (−8, 7) and (9, 4)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are is-

So,

Compare individual terms on both the sides-

So,

Similarly,

So,

So the co-ordinate of third vertex

#### Page No 6.67:

#### Question 14:

Write the coordinates the reflections of points (3, 5) in X and Y -axes.

#### Answer:

We have to find the reflection of (3, 5) along *x-*axis and *y-*axis.

Reflection of any pointalong *x-*axis is

So reflection of (3, 5) along *x-*axis is

Similarly, reflection of any pointalong *y-*axis is

So, reflection of (3, 5) along *y-*axis is

#### Page No 6.67:

#### Question 15:

If points Q and reflections of point P (−3, 4) in X and Y axes respectively, what is QR?

#### Answer:

We have to find the reflection of (−3, 4) along *x-*axis and *y-*axis.

Reflection of any pointalong *x-*axis is

So reflection of (−3, 4) along *x-*axis is

Similarly, reflection of any pointalong *y-*axis is

So, reflection of (−3, 4) along *y-*axis is

Therefore,

#### Page No 6.67:

#### Question 16:

Write the formula for the area of the triangle having its vertices at (x_{1}, y_{1}), (x2, y_{2}) and (x_{3}, y_{3}).

#### Answer:

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

The area ‘A’ encompassed by three points, and is also given by the formula,

#### Page No 6.67:

#### Question 17:

Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).

#### Answer:

The condition for co linearity of three points, and is that the area enclosed by them should be equal to 0.

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Thus for the three points to be collinear we need to have,

The area ‘A’ encompassed by three points, and is also given by the formula,

Thus for the three points to be collinear we can also have,

#### Page No 6.67:

#### Question 18:

Find the values of x for which the distance between the point P(2, −3), and Q (x, 5) is 10.

#### Answer:

It is given that distance between P (2,−3) and is 10.

In general, the distance between A and B is given by,

So,

On further simplification,

#### Page No 6.67:

#### Question 19:

Write the ratio in which the line segment doining the points A (3, −6), and B (5, 3) is divided by X-axis.

#### Answer:

Let P be the point of intersection of *x-*axis with the line segment joining A (3,−6) and B (5, 3) which divides the line segment AB in the ratio.

Now we will use section formula as,

Now equate the y component on both the sides,

On further simplification,

So *x-*axis divides AB in the ratio 2:1.

#### Page No 6.67:

#### Question 20:

Find the distance between the points $\left(-\frac{8}{5},2\right)$ and $\left(\frac{2}{5},2\right)$

#### Answer:

We have to find the distance between and .

In general, the distance between A and B is given by,

So,

#### Page No 6.67:

#### Question 22:

What is the distance between the points *A* (*c*, 0) and *B* (0, −*c*)?

#### Answer:

We have to find the distance between and .

In general, the distance between A and B is given by,

So,

#### Page No 6.67:

#### Question 23:

If *P* (2, 6) is the mid-point of the line segment joining *A* (6, 5) and *B* (4, *y*), find *y.*

#### Answer:

It is given that mid-point of line segment joining A (6, 5) and B (4, *y*) is P (2, 6)

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the y component to get,

So,

#### Page No 6.67:

#### Question 24:

If the distance between the points (3, 0) and (0, *y*) is 5 units and *y* is positive. then what is the value of *y*?

#### Answer:

It is given that distance between P (3, 0) and is 5.

In general, the distance between A and B is given by,

So,

On further simplification,

We will neglect the negative value. So,

#### Page No 6.67:

#### Question 25:

If *P* (*x*, 6) is the mid-point of the line segment joining *A* (6, 5) and *B* (4, *y*), find *y*.

#### Answer:

It is given that mid-point of line segment joining A (6, 5) and B (4, *y*) is

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the y component to get,

So,

#### Page No 6.68:

#### Question 26:

If *P* (2, *p*) is the mid-point of the line segment joining the points *A* (6, −5) and *B* (−2, 11). find the value of *p*.

#### Answer:

It is given that mid-point of line segment joining A (6,−5) and B (−2, 11) is

In general to find the mid-point of two pointsand we use section formula as,

So,

Now equate the *y* component to get,