NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry
Share with your friends Share Share Share twitter

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry are provided here with simple step-by-step explanations. These solutions for Coordinate Geometry are extremely popular among class 10 students for Maths Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 10 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 10 Maths are prepared by experts and are 100% accurate.

Page No 78:

Question 1:

Choose the correct answer from the given four options:
The distance of the point P (2, 3) from the x-axis is
(A) 2
(B) 3
(C) 1
(D) 5

Answer:

We know that, if $(x, y)$ is any point on the cartesian plane.

Then, $x$ = Perpendicular distance from Y-axis and $y$ = Perpendicular distance from X-axis.
∴ Distance of the point $P (2, 3)$ from the X-axis = ordinate of the point $P (2, 3)$ = 3.

Hence, the correct answer is option B.

Page No 78:

Question 2:

Choose the correct answer from the given four options:
The distance between the points A (0, 6) and B (0, –2) is
(A) 6
(B) 8
(C) 4
(D) 2

Answer:

The distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ .
Now, here $x_{1} = 0$ , $y_{1} = 6$ and $x_{2} = 0$ , $y_{2} = - 2$
∴ Distance between $A (0, 6)$ and $B (0, - 2)$ , AB = $\sqrt{{(0 - 0)}^{2} + {(- 2 - 6)}^{2}}$
$\Rightarrow$ AB = $\sqrt{0 + {(- 8)}^{2}} = \sqrt{8^{2}} = 8$

Hence, the correct answer is option B.

Page No 78:

Question 3:

Choose the correct answer from the given four options:
The distance of the point P (– 6, 8) from the origin is
(A) 8
(B) $2 \sqrt{7}$
(C) 10
(D) 6

Answer:

We know distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{{(x_{2} - x_{1_{}})}^{2} + {(y_{2} - y_{1})}^{2}}$ .
Here, $x_{1} = - 6$ , $y_{1} = 8$ and $x_{2} = 0$ , $y_{2} = 0$
∴ Distance between the points $P (- 6, 8)$ and origin i.e. $O (0, 0)$ ,
$\begin{array}{rcl} PO & = & \sqrt{[0 - (- 6))]^{2} + {(0 - 8)}^{2}} \\ = & \sqrt{{(6)}^{2} + {(- 8)}^{2}} \\ = & \sqrt{36 + 64} \\ = & \sqrt{100} \\ = & 10 \end{array}$

Thus, the distance between the two given points is 10.
Hence, the correct answer is option C.

Page No 78:

Question 4:

Choose the correct answer from the given four options:
The distance between the points (0, 5) and (–5, 0) is
(A) 5
(B) $5 \sqrt{2}$
(C) $2 \sqrt{5}$
(D) 10

Answer:

We know distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{{(x_{2} - x_{1_{}})}^{2} + {(y_{2} - y_{1})}^{2}}$
Here, $x_{1} = 0$ , $y_{1} = 5$ and $x_{2} = - 5$ , $y_{2} = 0$
∴ Distance between the points $(0, 5)$ and $(- 5, 0)$ = $\sqrt{{(- 5 - 0)}^{2} + {(0 - 5)}^{2}} = \sqrt{{(- 5)}^{2} + {(- 5)}^{2}}$
$= \sqrt{25 + 25}$ = $\sqrt{50} = 5 \sqrt{2}$
Thus, distance between the two given points is $5 \sqrt{2}$ .
Hence, the correct answer is option B.

Page No 78:

Question 5:

Choose the correct answer from the given four options:
AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(A) 5
(B) 3
(C) $\sqrt{34}$
(D) 4

Answer:

We know the distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{{(x_{2} - x_{1_{}})}^{2} + {(y_{2} - y_{1})}^{2}}$ .

Now, length of the diagonal AB = Distance between the points $A (0, 3)$ and $B (5, 0)$ .
Here, $x_{1} = 0$ , $y_{1} = 3$ and $x_{2} = 5$ , $y_{2} = 0$

∴ Distance between the points $A (0, 3)$ and $B (5, 0)$ , AB = $\sqrt{{(5 - 0)}^{2} + {(0 - 3)}^{2}}$
$= \sqrt{25 + 9} = \sqrt{34}$
Thus, the required length of diagonal is $\sqrt{34}$ .
Hence, the correct answer is option C.

Page No 78:

Question 6:

Choose the correct answer from the given four options:
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A) 5
(B) 12
(C) 11
(D) $7 + \sqrt{5}$

Answer:

Firstly we will determine the length of each side using the distance formula $\sqrt{{(x_{2} - x_{1_{}})}^{2} + {(y_{2} - y_{1})}^{2}}$ , further adding all the three sides of triangle to get the perimeter.

We have the vertices of a triangle are A $(0, 4)$ , O $(0, 0)$ and B $(3, 0)$ .

Now, perimeter of â–³AOB
= Sum of the length of all its sides
= OA + OB + AB
$= \sqrt{{(0 - 0)}^{2} + {(0 - 4)}^{2}} + \sqrt{{(3 - 0)}^{2} + {(0 - 0)}^{2}}$ $+ \sqrt{{(3 - 0)}^{2} + {(0 - 4)}^{2}}$
$= \sqrt{0 + 16} + \sqrt{9 + 0} + \sqrt{3^{2} + 4^{2}}$ $= 4 + 3 + \sqrt{9 + 16}$
$= 7 + \sqrt{25} = 7 + 5 = 12$
Thus, the perimeter of the triangle is 12.

Hence, the correct answer is option B.

Page No 78:

Question 7:

Choose the correct answer from the given four options:
The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(A) 14
(B) 28
(C) 8
(D) 6

Answer:

The area of triangle, whose vertices are $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ is given by $\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})|$ , using this formula we can find the area of given triangle.

Now, $x_{1} = 3$ , $y_{1} = 0$ , $x_{2} = 7$ , $y_{2} = 0$ , $x_{3} = 8$ and $y_{3} = 4$
∴ Area of $∆$ ABC
$= \frac{1}{2} |3 (0 - 4) + 7 (4 - 0) + 8 (0 - 0)| = \frac{1}{2} |(- 12 + 28 + 0)| = \frac{1}{2} |16| = 8$

Thus, the required area of $∆$ ABC is 8.
Hence, the correct answer is option C.

Page No 78:

Question 8:

Choose the correct answer from the given four options:
The points (–4, 0), (4, 0), (0, 3) are the vertices of a
(A) right triangle
(B) isosceles triangle
(C) equilateral triangle
(D) scalene triangle

Answer:

Let $A (- 4, 0)$ , $B (4, 0)$ , $C (0, 3)$ are the given vertices.

Now, distance between $A (- 4, 0)$ and $B (4, 0)$ , AB = $\sqrt{{[4 - (- 4)]}^{2} + {(0 - 0)}^{2}}$
$= \sqrt{{(4 + 4)}^{2}} = \sqrt{8^{2}} = 8$
Distance between $B (4, 0)$ and $C (0, 3)$ , BC = $\sqrt{{(0 - 4)}^{2} + {(3 - 0)}^{2}}$
$= \sqrt{16 + 9} = \sqrt{25} = 5$
Distance between $A (- 4, 0)$ and $C (0, 3)$ , AC = $\sqrt{{[0 - (- 4)]}^{2} + {(3 - 0)}^{2}}$
$= \sqrt{16 + 9} = \sqrt{25} = 5$

Since BC = AC, hence we conclude $△$ ABC is an isosceles triangle.
Hence, the correct answer is option B.

Page No 79:

Question 9:

Choose the correct answer from the given four options:
The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the
(A) I quadrant
(B) II quadrant
(C) III quadrant
(D) IV quadrant

Answer:

If $P (x, y)$ divides the line segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m : n$ , then
$x = \frac{m x_{2} + n x_{1}}{m + n}$ and $y = \frac{m y_{2} + n y_{1}}{m + n}$

Given, $x_{1} = 7$ , $y_{1} = - 6$ , $x_{2} = 3$ , $y_{2} = 4$ , $m = 1$ and $n = 2$
∴ $x = \frac{1 (3) + 2 (7)}{1 + 2}$ , $y = \frac{1 (4) + 2 (- 6)}{1 + 2}$ [by section formula]
$\Rightarrow x = \frac{3 + 14}{3}$ , $y = \frac{4 - 12}{3}$
$\Rightarrow x = \frac{17}{3}$ , $y = - \frac{8}{3}$
So, $P (x, y)$ $= (\frac{17}{3}, - \frac{8}{3})$ lies in IV quadrant.
Hence, the correct answer is option D.

Page No 79:

Question 10:

Choose the correct answer from the given four options:
The point which lies on the perpendicular bisector of the line segment joining the points A (–2, –5) and B (2, 5) is
(A) (0, 0)
(B) (0, 2)
(C) (2, 0)
(D) (–2, 0)

Answer:

We know that, the perpendicular bisector of the any line segment divides the line segment into two equal parts i.e. the perpendicular bisector of the line segment always passes through the mid-point of the line segment.

Since, mid-point of any line segment which passes through the points $(x_{1}, y_{1})$ and $(x_{2,} y_{2})$ is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

∴ Mid-point of the line segment joining the points $A (- 2, - 5)$ and $B (2, 5)$ is $(\frac{- 2 + 2}{2}, \frac{- 5 + 5}{2}) = (0, 0)$ .

Hence, the correct answer is option A.

Page No 79:

Question 11:

Choose the correct answer from the given four options:
The fourth vertex D of a parallelogram ABCD whose three vertices are A (–2, 3), B (6, 7) and C (8, 3) is
(A) (0, 1)
(B) (0, –1)
(C) (–1, 0)
(D) (1, 0)

Answer:

Let the fourth vertex of parallelogram, $D = (x_{4}, y_{4})$ and L, M be the middle points of AC and BD, respectively.
Since, mid-point of any line segment which passes through the points $(x_{1}, y_{1})$ and $(x_{2,} y_{2})$ $= (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
Then, L $= (\frac{- 2 + 8}{2}, \frac{3 + 3}{2}) = (3, 3)$ and M $= (\frac{6 + x_{4}}{2}, \frac{7 + y_{4}}{2})$
Now, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other.
Hence, L and M are the same points.

∴ $3 = \frac{6 + x_{4}}{2}$ and $3 = \frac{7 + y_{4}}{2}$
$\Rightarrow$ $6 = 6 + x_{4}$ and $6 = 7 + y_{4}$
$\Rightarrow x_{4} = 0$ and $y_{4} = 6 - 7$
$\Rightarrow x_{4} = 0$ and $y_{4} = - 1$

Hence, the correct answer is option B.

Page No 79:

Question 12:

Choose the correct answer from the given four options:
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(A) $AP = \frac{1}{3} AB$
(B) AP = PB
(C) $PB = \frac{1}{3} AB$
(D) $AP = \frac{1}{2} AB$

Answer:

Given, point $P (2, 1)$ lies on the line segment joining the points $A (4, 2)$ and $B (8, 4)$ as shown in figure below:

If $A (x, y)$ divides the line segment joining $P (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m : n$ , then
$x = \frac{m x_{2} + n x_{1}}{m + n}$ and $y = \frac{m y_{2} + n y_{1}}{m + n}$

Given, $x_{1} = 2$ , $y_{1} = 1$ , $x_{2} = 8$ , $y_{2} = 4$ , $x = 4$ and $y = 2$
Let AP : AB = m : n
∴ $4 = \frac{m (8) + n (2)}{m + n}$ [by section formula]
$\Rightarrow 4 (m + n) = 8 m + 2 n \Rightarrow 4 m + 4 n = 8 m + 2 n \Rightarrow 4 n - 2 n = 8 m - 4 m \Rightarrow n = 2 m$
Now, AP : AB = m : n, thus we can say AB = 2AP
∴ AP = $\frac{1}{2}$ AB
Hence, the correct answer is option D.

Page No 79:

Question 13:

Choose the correct answer from the given four options:
If $P (\frac{a}{3}, 4)$ is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is
(A) – 4
(B) – 12
(C) 12
(D) – 6

Answer:

Given, $P (\frac{a}{3}, 4)$ is the midpoint of the line segment joining the points Q (– 6, 5) and R (– 2, 3) as shown below:

Now, mid-point of any line segment which passes through the points $(x_{1}, y_{1})$ and $(x_{2,} y_{2})$ is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .
∴ Mid-point of QR = $P (\frac{- 6 - 2}{2}, \frac{5 + 3}{2}) = P (- 4, 4)$
However, the mid-point $P (\frac{a}{3}, 4)$ is given
$\Rightarrow (\frac{a}{3}, 4) = (- 4, 4)$
Comparing the coordinates, we get
$\frac{a}{3} = - 4$
$\Rightarrow a = - 12$
Thus, the required value of a is −12.

Hence, the correct answer is option B.

Page No 79:

Question 14:

Choose the correct answer from the given four options:
The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at
(A) (0, 13)
(B) (0, –13)
(C) (0, 12)
(D) (13, 0)

Answer:

Firstly, plot the points of the segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB i.e. perpendicular bisector of the line segment AB passes through the mid-point of AB.

∴ Mid-point of AB = $(\frac{1 + 4}{2}, \frac{5 + 6}{2})$
$\Rightarrow P = (\frac{5}{2}, \frac{11}{2})$

Now, we draw a straight line on paper that passes through the mid-point P. We see that the perpendicular bisector cuts the Y-axis at the point $(0, 13)$ .
Thus, the required point is $(0, 13)$ .
Hence, the correct answer is option A.

Page No 79:

Question 15:

Choose the correct answer from the given four options:
The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the Fig. 7.1 is

(A) (x, y)

(B) (y, x)

(C) $\frac{x}{2}, \frac{y}{2}$

(D) $\frac{y}{2}, \frac{x}{2}$

Answer:

Let the coordinate of the point which is equidistant from the three vertices $O (0, 0)$ , $A (0, 2 y)$ and $B (2 x, 0)$ is $P (h, k)$ .
Then, PO = PA = PB
$\Rightarrow {(PO)}^{2} = {(PA)}^{2} = {(PB)}^{2}$

Thus, by the distance formula
$\sqrt{{(h - 0)}^{2} + {(k - 0)}^{2}} = \sqrt{{(h - 0)}^{2} + {(k - 2 y)}^{2}} \begin{matrix} \end{matrix}$ $= \sqrt{{(h - 2 x)}^{2} + {(k - 0)}^{2}}$
$\Rightarrow h^{2} + k^{2} = h^{2} + {(k - 2 y)}^{2} = {(h - 2 x)}^{2} + k^{2}$
Now, taking first two term, we get
$h^{2} + k^{2} = h^{2} + {(k - 2 y)}^{2} \Rightarrow k^{2} = k^{2} + 4 y^{2} - 4 y k \Rightarrow 4 y (y - k) = 0$
$\Rightarrow y = k$ , since $y \neq 0$
Now, taking first and third term, we get
$h^{2} + k^{2} = {(h - 2 x)}^{2} + k^{2} \Rightarrow h^{2} = h^{2} + 4 x^{2} - 4 x h \Rightarrow 4 x (x - h) = 0$
$\Rightarrow x = h$ , since $x \neq 0$

Thus, the required point is $(x, y)$ .
Hence, the correct answer is option A.

Page No 79:

Question 16:

Choose the correct answer from the given four options:
A circle drawn with origin as the centre passes through $(\frac{13}{2}, 0)$ . The point which does not lie in the interior of the circle is
(A) $(\frac{- 3}{4}, 1)$
(B) $(2, \frac{7}{3})$
(C) $(5, \frac{- 1}{2})$
(D) $(- 6, \frac{5}{2})$

Answer:

Given, coordinates of the centre of circle = $(0, 0)$ and passes through the point $(\frac{13}{2}, 0)$ .
∴ Radius of circle = Distance between $(0, 0)$ and $(\frac{13}{2}, 0)$
$= \sqrt{{(\frac{13}{2} - 0)}^{2} + {(0 - 0)}^{2}} = \sqrt{{(\frac{13}{2})}^{2}} = \frac{13}{2} = 6.5$

Now, a point will lie inside the circle if the distance between the given point and centre of circle is less than radius

Distance between $(0, 0)$ and $(- 6, \frac{5}{2})$ $= \sqrt{{(- 6 - 0)}^{2} + {(\frac{5}{2} - 0)}^{2}}$
$= \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{144 + 25}{4}}$
$= \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5$

Thus, distance between given point and centre of circle is equal to the radius. So, the point $(- 6, \frac{5}{2})$ does not lie in the interior of the circle.
Hence, the correct answer is option D.

Page No 80:

Question 17:

Choose the correct answer from the given four options:
A line intersects the y-axis and x-axis at the points 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:

Let the coordinates of P and Q are $(0, y)$ , $(x, 0)$ , respectively.
So, the mid-point of $P (0, y)$ and $Q (x, 0)$ is $M (\frac{0 + x}{2}, \frac{y + 0}{2})$

However, it is given that mid-point of $P Q$ is $(2, - 5)$ .
$\Rightarrow 2 = \frac{x + 0}{2}$ and $- 5 = \frac{y + 0}{2}$
∴ $x = 4$ and $y = - 10$

So, the coordinates of P and Q are $(0, - 10)$ and $(4, 0)$
Hence, the correct answer is option D.

Page No 80:

Question 18:

Choose the correct answer from the given four options:
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(A) (a + b + c)²
(B) 0
(C) a + b + c
(D) abc

Answer:

Let the vertices of a triangle are,
$A (x_{1}, y_{1}) = (a, b + c) B (x_{2}, y_{2}) = (b, c + a) C (x_{3}, y_{3}) = (c, a + b)$

âˆµ Area of $△$ ABC $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$= \frac{1}{2} [a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a)]$
$= \frac{1}{2} [(a c - a b + a b - b c + b c - a c)] = \frac{1}{2} (0) = 0$

Hence, the correct answer is option B.

Page No 80:

Question 19:

Choose the correct answer from the given four options:
If the distance between the points (4, p) and (1, 0) is 5, then the value of p is
(A) 4 only
(B) ± 4
(C) – 4 only
(D) 0

Answer:

The distance between the points $(4, p)$ and $(1, 0)$ is 5.
i.e. distance $= \sqrt{{(1 - 4)}^{2} + {(0 - p)}^{2}} = 5$
$\Rightarrow \sqrt{{(- 3)}^{2} + p^{2}} = 5 \Rightarrow \sqrt{9 + p^{2}} = 5$
Squaring both sides, we get
$9 + p^{2} = 25$
$p^{2} = 16 \Rightarrow p = \pm 4$

Hence, the correct answer is option B.

Page No 80:

Question 20:

Choose the correct answer from the given four options:
If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then
(A) a = b
(B) a = 2b
(C) 2a = b
(D) a = – b

Answer:

Let the given points are
$A (x_{1}, y_{1}) = (1, 2) O (x_{2}, y_{2}) = (0, 0) C (x_{3}, y_{3}) = (a, b)$

∴ Area of $△$ ABC $= \frac{1}{2} [1 (0 - b) + 0 (b - 2) + a (2 - 0)]$
$= \frac{1}{2} (- b + 0 + 2 a) = \frac{1}{2} (2 a - b)$

Since, the points are collinear, then area of $△$ ABC should be equal to zero.
$\Rightarrow \frac{1}{2} (2 a - b) = 0 \Rightarrow 2 a - b = 0 \Rightarrow 2 a = b$

Hence, the correct answer is option C.

Page No 80:

Question 1:

State whether the following statements are true or false. Justify your answer.
ΔABC with vertices A (–2, 0), B (2, 0) and C (0, 2) is similar to ΔDEF with vertices D (–4, 0) E (4, 0) and F (0, 4).

Answer:

In $△$ ABC,
Distance between AB= $\sqrt{{[2 - (- 2)]}^{2} + {(0 - 0)}^{2}} = 4$
Distance between BC $= \sqrt{{(0 - 2)}^{2} + {(2 - 0)}^{2}} = 2 \sqrt{2}$
Distance between CA $= \sqrt{{[0 - (- 2)]}^{2} + {(2 - 0)}^{2}} = 2 \sqrt{2}$

Now, in $△$ DEF,
Distance between DF $= \sqrt{{[0 - (- 4)]}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2}$
Distance between EF $= \sqrt{{(0 - 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2}$
Distance between DE $= \sqrt{{[4 - (- 4)]}^{2} + {(0 - 0)}^{2}} = 8$

Thus, we can say
$\frac{AB}{ED} = \frac{4}{8} = \frac{1}{2}$ , $\frac{AC}{DF} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2}$ and $\frac{BC}{EF} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{1}{2}$
$\Rightarrow \frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$

Thus, sides are proportional, So we conclude both the triangles are similar.

Page No 81:

Question 2:

State whether the following statements are true or false. Justify your answer.
Point P (– 4, 2) lies on the line segment joining the points A (–4, 6) and B (–4, –6).

Answer:

We plot all the points P (– 4, 2), A (–4, 6) and B (–4, –6) on the graph paper.

From the figure, we conclude point P (– 4, 2) lies on the line segment joining the points A (–4, 6) and B (–4, –6).

Page No 81:

Question 3:

State whether the following statements are true or false. Justify your answer.
The points (0, 5), (0, –9) and (3, 6) are collinear.

Answer:

Here $x_{1} = 0$ , $x_{2} = 0$ , $x_{3} = 3$ and $y_{1} = 5$ , $y_{2} = - 9$ , $y_{3} = 6$

âˆµ Area of triangle $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
Thus,
Area $= \frac{1}{2} [0 (- 9 - 6) + 0 (6 - 5) + 3 (5 + 9)]$
$= \frac{1}{2} (0 + 0 + 3 \times 14) = 21 \neq 0$
As we know, the area of triangle formed by the points (0, 5), (0, –9) and (3, 6) is zero.
Thus, we conclude points are collinear.

Page No 81:

Question 4:

State whether the following statements are true or false. Justify your answer.
Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A (–1, 1) and B (3, 3).

Answer:

We know that, the point lie on perpendicular bisector of the line segment joining the two points is equidistant from these two points.
Thus, if point P(0, 2) is the perpendicular bisector of the points A (–1, 1) and B (3, 3) then PA = PB
$\begin{array}{rcl} PA & = & \sqrt{{(- 1 - 0)}^{2} + {(1 - 2)}^{2}} \\ = & \sqrt{{(- 1)}^{2} + {(- 1)}^{2}} \\ = & \sqrt{1 + 1} \\ = & \sqrt{2} \end{array}$

$\begin{array}{rcl} PB & = & \sqrt{{(3 - 0)}^{2} + {(3 - 2)}^{2}} \\ = & \sqrt{3^{2} + 1^{2}} \\ = & \sqrt{10} \end{array}$

Since, PA $\neq$ PB
Thus, we conclude the point P(0, 2) does not lie on the perpendicular bisector of AB.

Page No 81:

Question 5:

State whether the following statements are true or false. Justify your answer.
Points A(3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.

Answer:

Let $A (3, 1) = (x_{1}, y_{1})$ , $B (12, - 2) = (x_{2}, y_{2})$ and $C (0, 2) = (x_{3}, y_{3})$
∴ Area of $△$ ABC $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$= \frac{1}{2} [3 (- 2 - 2) + 12 (2 - 1) + 0 \{1 - (- 2)\}] = \frac{1}{3} [3 (- 4) + 12 (1) + 0] = \frac{1}{3} (- 12 + 12) = 0$

Thus, the area of $△$ ABC is 0.

Hence, the points A(3, 1), B(12, –2) and C(0, 2) are collinear.
So the points cannot be the vertices of a triangle.

Page No 81:

Question 6:

State whether the following statements are true or false. Justify your answer.
Points A(4, 3), B (6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.

Answer:

In parallelogram, opposite sides are equal, here we need to find the length of sides AB, BC, CD and DA using distance formula
if the pair of sides are equal then we can say it is a parallelogram.

Distance between A(4, 3) and B (6, 4), AB $= \sqrt{{(6 - 4)}^{2} + {(4 - 3)}^{2}}$ $= \sqrt{2^{2} + 1^{2}} = \sqrt{5}$
Distance between B (6, 4) and C(5, –6), BC $= \sqrt{{(5 - 6)}^{2} + {(- 6 - 4)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 10)}^{2}} = \sqrt{101}$
Distance between C(5, –6) and D(–3, 5), CD $= \sqrt{{(- 3 - 5)}^{2} + {(5 + 6)}^{2}} = \sqrt{{(- 8)}^{2} + 11^{2}} = \sqrt{185}$
Distance between D(–3, 5) and A(4, 3), DA $= \sqrt{{(4 + 3)}^{2} + {(3 - 5)}^{2}} = \sqrt{7^{2} + {(- 2)}^{2}} = \sqrt{53}$

Now, no pair of sides are equal , thus we conclude given vertices are not the vertices of parallelogram.

Page No 81:

Question 7:

State whether the following statements are true or false. Justify your answer.
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.

Answer:

From the given information, first we find the distance between the centre of circle and the given point

Distance between origin O(0, 0) and P(5, 0),
OP $= \sqrt{{(5 - 0)}^{2} + {(0 - 0)}^{2}}$ $= \sqrt{5^{2} + 0^{2}} = 5$
Thus, the radius of the circle is 5.

Now, we know that if the distance of any point from the centre is more than radius , then the point lie outside the circle.

Distance between origin O(0, 0) and Q(6, 8),
OQ $= \sqrt{{(6 - 0)}^{2} + {(8 - 0)}^{2}} = \sqrt{6^{2} + 8^{2}} = 10$
Here, we see that, $OQ > OP$ .

Hence, it is true that point Q(6, 8), lies outside the circle.

Page No 81:

Question 8:

State whether the following statements are true or false. Justify your answer.
The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, –4).

Answer:

If A(2, 7) lies on perpendicular bisector of P(6, 5) and Q(0, –4), then AP = AQ

∴ AP $= \sqrt{{(6 - 2)}^{2} + {(5 - 7)}^{2}} = \sqrt{4^{2} + {(- 2)}^{2}} = \sqrt{20}$ and,
AQ $= \sqrt{{(0 - 2)}^{2} + {(- 4 - 7)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 11)}^{2}} = \sqrt{125}$

Thus, we conclude AP $\neq$ AQ.
Hence, A does not lie on the perpendicular bisector of PQ.

Page No 81:

Question 9:

State whether the following statements are true or false. Justify your answer.
Point P(5, –3) is one of the two points of trisection of the line segment joining the points A(7, –2) and B(1, –5).

Answer:

Let P(5, –3) divides the line segment joining the points A(7, –2) and B(1, –5) in the ratio k : 1 internally.
By section formula, the coordinate of point P will be
$(\frac{k (1) + (1) 7}{k + 1}, \frac{k (- 5) + 1 (- 2)}{k + 1})$
$\Rightarrow (\frac{k + 7}{k + 1}, \frac{- 5 k - 2}{k + 1})$
$\Rightarrow (5, - 3) = (\frac{k + 7}{k + 1}, \frac{- 5 k - 2}{k + 1}) \Rightarrow \frac{k + 7}{k + 1} = 5 \Rightarrow k + 7 = 5 k + 5 \Rightarrow - 4 k = - 2 \Rightarrow k = \frac{1}{2}$

So the point P divides the line segment AB in ratio 1 : 2
Hence, point P is one of the two points of trisection.

Page No 81:

Question 10:

State whether the following statements are true or false. Justify your answer.
Points A(–6, 10), B(–4, 6) and C(3, –8) are collinear such that $AB = \frac{2}{9} AC$ .

Answer:

If the area of triangle formed by the points A(–6, 10), B(–4, 6) and C(3, –8) is zero, then the points are collinear.

Here, $x_{1} = - 6$ , $x_{2} = - 4$ , $x_{3} = 3$ and $y_{1} = 10$ , $y_{2} = 6$ , $y_{3} = - 8$
∴ Area of $△$ ABC $= \frac{1}{2} [- 6 \{6 - (- 8)\} + (- 4) (- 8 - 10) + 3 (10 - 6)]$
$= \frac{1}{2} (- 84 + 72 + 12) = 0$
So, given points are collinear.

Now, Distance between A(–6, 10) and B(–4, 6), AB $= \sqrt{{(- 4 + 6)}^{2} + {(6 - 10)}^{2}} = \sqrt{20} = 2 \sqrt{5}$
and, distance between A(–6, 10) and C(3, –8), AC $= \sqrt{{(3 + 6)}^{2} + {(- 8 - 10)}^{2}} = \sqrt{405} = 9 \sqrt{5}$

Thus, from above we conclude AB $= \frac{2}{9}$ AC.

Page No 81:

Question 11:

State whether the following statements are true or false. Justify your answer.
The point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).

Answer:

If the distance between the centre and any point is equal to the radius, then we say that point lie on the circle.

Now, distance between P(–2, 4) and centre C(3, 5) $= \sqrt{{(3 + 2)}^{2} + {(5 - 4)}^{2}} = \sqrt{26}$
which is not equal to the radius of the circle.
Hence, the point P(–2, 4) does not lie on the circle.

Page No 81:

Question 12:

State whether the following statements are true or false. Justify your answer.
The points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) in that order form a rectangle.

Answer:

Distance between A(–1, –2) and B(4, 3), AB $= \sqrt{{(4 + 1)}^{2} + {(3 + 2)}^{2}} = \sqrt{50} = 5 \sqrt{2}$
Distance between C(2, 5) and D(–3, 0), CD $= \sqrt{{(- 3 - 2)}^{2} + {(0 - 5)}^{2}} = \sqrt{50} = 5 \sqrt{2}$
Distance between A(–1, –2) and D(–3, 0), AD $= \sqrt{{(- 3 + 1)}^{2} + {(0 + 2)}^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$
Distance between B(4, 3) and C(2, 5), BC $= \sqrt{{(4 - 2)}^{2} + {(3 - 5)}^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$

We know that, in a rectangle, opposite sides and diagonals are equal and bisect each other.
Thus, AB = CD and AD = BC.

Now, for diagonals
Distance between A(–1, –2) and C(2, 5), AC $= \sqrt{{(2 + 1)}^{2} + {(5 + 2)}^{2}} = \sqrt{9 + 49} = \sqrt{58}$
Distance between D(–3, 0) and B(4, 3), DB $= \sqrt{{(4 + 3)}^{2} + {(3 - 0)}^{2}} = \sqrt{49 + 9} = \sqrt{58}$

Now, diagonals AC and BD are equal.
Hence, the points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) form a rectangle.

Page No 83:

Question 1:

Name the type of triangle formed by the points A(–5, 6), B(–4, –2) and C(7, 5).

Answer:

First we determine the length of all the three sides and check whatever the condition of triangle is satisfied by these sides.

Thus, using distance formula between two points,
AB $= \sqrt{{(- 4 + 5)}^{2} + {(- 2 - 6)}^{2}} = \sqrt{1 + 64} = \sqrt{65}$
BC $= \sqrt{{(7 + 4)}^{2} + {(5 + 2)}^{2}} = \sqrt{121 + 49} = \sqrt{170}$
AC $= \sqrt{{(- 5 - 7)}^{2} + {(6 - 5)}^{2}} = \sqrt{144 + 1} = \sqrt{145}$

We conclude AB $\neq$ BC $\neq$ AC, and does not satisfy the condition of pythagoras.
Hence, the required triangle is scalene.

Page No 83:

Question 2:

Find the points on the x-axis which are at a distance of $2 \sqrt{5}$ from the point (7, –4). How many such points are there?

Answer:

We know that, every point on the X-axis is in the form of $(x, 0)$ . Let $P (x, 0)$ the point on the X-axis have $2 \sqrt{5}$ distance from the point $Q (7, - 4)$
By given condition, PQ $= 2 \sqrt{5}$

$\Rightarrow {(PQ)}^{2} = 4 \times 5$
$\Rightarrow {(x - 7)}^{2} + {(0 + 4)}^{2} = 20 \Rightarrow x^{2} + 49 - 14 x + 16 = 20 \Rightarrow x^{2} - 14 x + 65 - 20 = 0 \Rightarrow x^{2} - 14 x + 45 = 0 \Rightarrow x^{2} - 9 x - 5 x + 45 = 0 \Rightarrow x (x - 9) - 5 (x - 9) = 0 \Rightarrow (x - 9) (x - 5) = 0$

∴ $x = 5, 9$
Hence, we conclude two points lie on the X-axis, which are $(5, 0)$ and $(9, 0)$ , have $2 \sqrt{5}$ distance from the point $(7, - 4)$ .

Page No 83:

Question 3:

What type of a quadrilateral do the points A(2, –2), B(7, 3), C(11, –1) and D(6, –6) taken in that order, form?

Answer:

To find the type of quadrilateral, the length of all four sides as well as two diagonals as shown below

Now, using distance formula we can find the sides of quadrilateral.
AB $= \sqrt{{(7 - 2)}^{2} + {(3 + 2)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2}$
BC $= \sqrt{{(11 - 7)}^{2} + {(- 1 - 3)}^{2}} = \sqrt{4^{2} + {(- 4)}^{2}} = \sqrt{32} = 4 \sqrt{2}$
CD $= \sqrt{{(6 - 11)}^{2} + {(- 6 + 1)}^{2}} = \sqrt{50} = 5 \sqrt{2}$
AD $= \sqrt{{(2 - 6)}^{2} + {(- 2 + 6)}^{2}} = \sqrt{32} = 4 \sqrt{2}$

Similarly, we can find the length of diagonals.
AC $= \sqrt{{(11 - 2)}^{2} + {(- 1 + 2)}^{2}} = \sqrt{81 + 1} = \sqrt{82}$
BD $= \sqrt{{(6 - 7)}^{2} + {(- 6 - 3)}^{2}} = \sqrt{1 + 81} = \sqrt{82}$

Thus, we conclude AB = CD and BC = AD.
Also, AC = BD, diagonals are equal.
Hence, the required quadrilateral is a rectangle.

Page No 83:

Question 4:

Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units.

Answer:

Distance between A(–3, –14) and B(a, –5), AB = 9
Now, using distance formula we get
AB $= \sqrt{{(a + 3)}^{2} + {(- 5 + 14)}^{2}}$
$\Rightarrow \sqrt{{(a + 3)}^{2} + 9^{2}} = 9$

Squaring both sides, we get
${(a + 3)}^{2} + 81 = 81 \Rightarrow {(a + 3)}^{2} = 0 \Rightarrow a = - 3$

Hence, the required value of a is −3.

Page No 83:

Question 5:

Find a point which is equidistant from the points A(–5, 4) and B(–1, 6)? How many such points are there?

Answer:

Let $P (h, k)$ be the point which is equidistant from the points A(–5, 4) and B(–1, 6).
∴ PA = PB
$\Rightarrow {(PA)}^{2} = {(PB)}^{2}$
$\Rightarrow {(- 5 - h)}^{2} + {(4 - k)}^{2} = {(- 1 - h)}^{2} + {(6 - k)}^{2} \Rightarrow 25 + h^{2} + 10 h + 16 + k^{2} - 8 k = 1 + h^{2} + 2 h + 36 + k^{2} - 12 k \Rightarrow 25 + 10 h + 16 - 8 k = 1 + 2 h + 36 - 12 k \Rightarrow 8 h + 4 k + 41 - 37 = 0 \Rightarrow 8 h + 4 k + 4 = 0 \Rightarrow 2 h + k + 1 = 0$

Now, the mid-point of AB $= (\frac{- 5 - 1}{2}, \frac{4 + 6}{2}) = (- 3, 5)$
Now point $(- 3, 5)$ should satisfy the equation $2 h + k + 1 = 0$
$\Rightarrow 2 h + k = 2 (- 3) + 5 = - 6 + 5 = - 1 \Rightarrow 2 h + k + 1 = 0$
So, we conclude the mid-point of AB satisfy the given equation
Hence, all the points which are solutions of the equation $2 h + k + 1 = 0$ , are equidistant from the points A and B.

Page No 84:

Question 6:

Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.

Answer:

The perpendicular bisector of the line segment AB bisects the segment AB i.e. perpendicular bisector of the line segment AB passes through the mid-point of AB.
∴ Mid-point of AB $= (\frac{- 5 + 4}{2}, \frac{- 2 - 2}{2})$
$\Rightarrow R = (- \frac{1}{2}, - 2)$

Now, we draw a straight line passing through the mid-point R. We see that the perpendicular bisector cuts X-axis at the point $Q (- \frac{1}{2}, 0)$ .
Hence, the required coordinates are $Q (- \frac{1}{2}, 0)$ .

To know the type of triangle formed by the points Q, A and B. We find the length of all three sides
Now, using distance formula we get
AB $= \sqrt{{(4 + 5)}^{2} + {(- 2 + 2)}^{2}} = \sqrt{9^{2} + 0} = 9$
BQ $= \sqrt{{(\frac{- 1}{2} - 4)}^{2} + {(0 + 2)}^{2}} = \sqrt{\frac{81}{4} + 4} = \sqrt{\frac{97}{4}} = \frac{\sqrt{97}}{2}$
QA $= \sqrt{{(- 5 + \frac{1}{2})}^{2} + {(- 2 - 0)}^{2}} = \sqrt{\frac{81}{4} + 4} = \frac{\sqrt{97}}{2}$

Thus, we conclude BQ = QA $\neq$ AB.
Hence, the triangle formed by the points Q, A and B is an isosceles.

Page No 84:

Question 7:

Find the value of m if the points (5, 1), (–2, –3) and (8, 2m) are collinear.

Answer:

Let $A (x_{1}, y_{1}) = (5, 1), B (x_{2}, y_{2}) = (- 2, - 3), C (x_{2}, y_{3}) = (8, 2 m)$
Since, the points are collinear, thus area of $△$ ABC = 0.
$\Rightarrow \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0 \Rightarrow \frac{1}{2} [5 (- 3 - 2 m) + (- 2) (2 m - 1) + 8 \{1 - (- 3)\}] = 0 \Rightarrow \frac{1}{2} (- 15 - 10 m - 4 m + 2 + 32) = 0 \Rightarrow \frac{1}{2} (- 14 m + 19) = 0 \Rightarrow m = \frac{19}{14}$

Hence, the required value of m is $\frac{19}{14}$ .

Page No 84:

Question 8:

If the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y), find the values of y. Also find distance PQ.

Answer:

Distance between two points (x₁, y₁) and (x₂, y₂) is given as: $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
Therefore,
AP = $\sqrt{{(3 - 2)}^{2} + {(8 - (- 4))}^{2}} = \sqrt{145}$
AQ = $\sqrt{{(- 10 - 2)}^{2} + {(y - (- 4))}^{2}} = \sqrt{144 + {(y + 4)}^{2}}$
PQ= $\sqrt{{(- 10 - 3)}^{2} + {(y - 8)}^{2}} = \sqrt{169 + {(y - 8)}^{2}}$
Now,
AP = AQ (Given)
∴ $\sqrt{145} = \sqrt{144 + {(y + 4)}^{2}}$
Squaring both sides, we get
145 = 144 + (y + 4)²
y + 4 = $\pm$ 1
y = −3 OR y = −5

Case-I: y = −3
PQ = $\sqrt{169 + {(- 3 - 8)}^{2}} = \sqrt{169 + 121} = \sqrt{290}$
Case-II: y = −5
PQ = $\sqrt{169 + {(- 5 - 8)}^{2}} = \sqrt{169 + 169} = 13 \sqrt{2}$

Hence, required value of PQ is $\sqrt{290}$ OR $13 \sqrt{2}$ .

Page No 84:

Question 9:

Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).

Answer:

Let (x₁, y₁) → (–8, 4), (x₂, y₂) → (–6, 6) and (x₃, y₃) → (–3, 9).
We know area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
$\begin{array}{rcl} △ & = & \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \\ = & \frac{1}{2} [- 8 (6 - 9) - 6 (9 - 4) - 3 (4 - 6)] \\ = & \frac{1}{2} (30 - 30) \\ = & 0 \end{array}$

Hence, the required area of the triangle is 0.

Page No 84:

Question 10:

In what ratio does the x-axis divide the line segment joining the points (–4, –6) and (–1, 7)? Find the coordinates of the point of division.

Answer:

Let the line segment joining the points (−4, −6) and (−1, 7) be divided by the point on x-axis (x, 0) in the ratio k : 1.
The coordinates of the point dividing the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m : n internally is given by $(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$ .

$∴ (\frac{k \times (- 1) + 1 \times (- 4)}{k + 1}, \frac{k \times 7 + 1 \times (- 6)}{k + 1}) = (\frac{- k - 4}{k + 1}, \frac{7 k - 6}{k + 1}) \Rightarrow 0 = \frac{7 k - 6}{k + 1} \Rightarrow k = \frac{6}{7}$
Therefore, the required ratio is 6 : 7.
Using section formula, we have:
$\frac{- k - 4}{k + 1} = \frac{\frac{- 6}{7} - 4}{\frac{6}{7} + 1} = \frac{- 6 - 28}{13} = \frac{- 34}{13}$

Hence, the coordinates of the point of division is $(\frac{- 34}{13}, 0)$ .

Page No 84:

Question 11:

Find the ratio in which the point $P (\frac{3}{4}, \frac{5}{12})$ divides the line segment joining the points $A (\frac{1}{2}, \frac{3}{2})$ and B(2, –5).

Answer:

Let the required ratio be k : 1.
Then, the coordinates of P are $(\frac{2 k + \frac{1}{2}}{k + 1}, \frac{- 5 k + \frac{3}{2}}{k + 1})$ .
$∴ \frac{2 k + \frac{1}{2}}{k + 1} = \frac{3}{4} \Rightarrow \frac{4 k + 1}{2 (k + 1)} = \frac{3}{4} \Rightarrow 16 k + 4 = 6 k + 6 \Rightarrow k = \frac{1}{5}$

Hence, the required ratio is 1 : 5.

Page No 84:

Question 12:

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

Answer:

Given: Point P divides AB in the ratio 3:1
By using section formula, we have
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n}$
$∴ \frac{3 (8 a) + 3 a + 1}{4} = 9 a - 2 \Rightarrow 27 a + 1 = 36 a - 8 \Rightarrow 9 a = 9 \Rightarrow a = 1$

Also,
$\frac{3 (5) + 1 (- 2)}{4} = - b \Rightarrow - 4 b = 13 \Rightarrow b = \frac{- 13}{4}$

Hence, the required values of a and b are 1 and $\frac{- 13}{4}$ .

Page No 84:

Question 13:

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

Answer:

Since, (a,b) is the mid-point of line segment AB
$(a, b) = (\frac{10 + k}{2}, \frac{- 6 + 4}{2}) (a, b) = (\frac{10 + k}{2}, - 1)$
Now, equating coordinates on both sides, we get
$a = \frac{10 + k}{2} . . . . . (1) and b = - 1 . . . . . (2)$

a − 2b = 18 (Given)
From (2), we have
a − 2(−1) = 18
$\Rightarrow$ a = 16

From (1), we have
$\frac{10 + k}{2} = 16 \Rightarrow k = 22$

Now, the distance between A(10, −6) and B(22, 4)
$\begin{array}{rcl} AB & = & \sqrt{{(22 - 10)}^{2} + {(4 - (- 6))}^{2}} \\ = & \sqrt{{(12)}^{2} + {(10)}^{2}} \\ = & \sqrt{244} \\ = & 2 \sqrt{61} \end{array}$

Hence, the required value of k is 22 and distance AB is $2 \sqrt{61}$ .

Page No 84:

Question 14:

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:

Distance between the centre C(2a, a – 7) and the point P(11, –9), which lie on the circle = Radius of circle
Radius of circle = $\sqrt{{(11 - 2 a)}^{2} + {(- 9 - a + 7)}^{2}}$ .....(1)
Length of diameter = $10 \sqrt{2}$ units (Given)
$∴ \sqrt{{(11 - 2 a)}^{2} + {(a + 2)}^{2}} = 5 \sqrt{2} (∵ Radius = \frac{Diameter}{2})$
Squaring on both sides, we get
${(11 - 2 a)}^{2} + {(a + 2)}^{2} = 50 \Rightarrow 121 + 4 a^{2} - 44 a + a^{2} + 4 + 4 a = 50 \Rightarrow 5 a^{2} + 125 - 40 a = 50 \Rightarrow 5 a^{2} + 75 - 40 a = 0 \Rightarrow a^{2} - 8 a + 15 = 0 \Rightarrow (a - 5) (a - 3) = 0 ∴ a = 3, 5$

Hence, the required values of a are 3 and 5.

Page No 84:

Question 15:

The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

Answer:

Given: Points A(3, 2) and B(5, 1) divides the line segment in the ratio 1 : 2
By section formula, we have
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n}$
$∴ x = \frac{1 \times 5 + 2 \times 3}{1 + 2}, y = \frac{1 \times 1 + 2 \times 2}{1 + 2} \Rightarrow (x, y) = (\frac{11}{3}, \frac{5}{3})$

Now, this above point lies on 3x – 18y + k = 0
$∴ 3 (\frac{11}{3}) - 18 (\frac{5}{3}) + k = 0 \Rightarrow 11 - 30 + k = 0 \Rightarrow k = 19$

Hence, the required value of k is 19.

Page No 84:

Question 16:

If $D (\frac{- 1}{2}, \frac{5}{2})$ , E (7, 3) and $F (\frac{7}{2}, \frac{7}{2})$ are the midpoints of sides of ΔABC, find the area of the ΔABC.

Answer:

Given: $D (\frac{- 1}{2}, \frac{5}{2})$ , E (7, 3) and $F (\frac{7}{2}, \frac{7}{2})$ are the midpoints of sides BC, CB and AB.
Let A → (x₁, y₁), B → (x₂, y₂) and C → (x₃, y₃) are the vertices of $△ ABC$
Since, $D (\frac{- 1}{2}, \frac{5}{2})$ is the mid-point of BC
$∴ \frac{x_{2} + x_{3}}{2} = \frac{- 1}{2}, \frac{y_{2} + y_{3}}{2} = \frac{5}{2} \Rightarrow x_{2} + x_{3} = - 1 . . . . . (1) and y_{2} + y_{3} = 5 . . . . . (2)$
Also, E (7, 3) is the mid-point of CA
$∴ \frac{x_{3} + x_{1}}{2} = 7 and \frac{y_{3} + y_{1}}{2} = 3 \Rightarrow x_{3} + x_{1} = 14 . . . . . (3) \Rightarrow y_{3} + y_{1} = 6 . . . . . (4)$
Similarly, $F (\frac{7}{2}, \frac{7}{2})$ is the mid-point of AB
$∴ \frac{x_{1} + x_{2}}{2} = \frac{7}{2} and \frac{y_{1} + y_{2}}{2} = \frac{7}{2} \Rightarrow x_{1} + x_{2} = 7 . . . . . (5) \Rightarrow y_{1} + y_{2} = 7 . . . . . (6)$
Adding equation(1), (3) and (5), we get
2(x₁ + x₂ + x₃) = 20
$\Rightarrow$ (x₁ + x₂ + x₃) = 10 .....(7)
Subtracting equations(1), (3) and (5) from (7), we get
x₁ = 11, x₂ = −4, x₃ = 3
Again, adding equations(2), (4) and (6), we get
2(y₁ + y₂ + y₃) = 18
$\Rightarrow$ (y₁ + y₂ + y₃) = 9 .....(8)
Subtracting equations(2), (4) and (6) from (8), we get
y₁ = 4, y₂ = 3, y₃ = 2
Now, area of Triangle ABC is given by
$\begin{array}{rcl} △ & = & \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \\ = & \frac{1}{2} [11 (3 - 2) + (- 4) (2 - 4) + 3 (4 - 3)] \\ = & \frac{1}{2} [11 + (- 4) (- 2) + 3 (1)] \\ = & 11 \end{array}$

Hence, the required area of the triangle is 11.

Page No 84:

Question 17:

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

Answer:

Given: A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ΔABC right angled at B
Now,
AB = $\sqrt{{(a - 2)}^{2} + {(5 - 9)}^{2}}$ (âˆµ Distance between two points (x₁, y₁) and (x₂, y₂) = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ )
= $\sqrt{a^{2} - 4 a + 20}$
BC = $\sqrt{{(a - 5)}^{2} + {(5 - 5)}^{2}} = a - 5$
CA = $\sqrt{{(9 - 5)}^{2} + {(2 - 5)}^{2}} = 5$
Here, (AC)² = (AB)² + (BC)²
$25 = (a^{2} - 4 a + 20) + {(a - 5)}^{2} 25 = 2 a^{2} - 14 a + 45 \Rightarrow 2 a^{2} - 14 a + 20 = 0 \Rightarrow a^{2} - 7 a + 10 = 0 \Rightarrow (a - 5) (a - 2) = 0 \Rightarrow a = 2 (a \neq 5, Here BC = 0)$

Now, coordinates of A, B and C are (2, 9), (2, 5) and (5, 5).
Area of triangle = $\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$\begin{array}{rcl} △ & = & \frac{1}{2} [2 (5 - 5) + 2 (5 - 9) + 5 (9 - 5)] \\ = & \frac{1}{2} [0 - 8 + 20] \\ = & 6 sq units \end{array}$

Hence, the area of $△ ABC$ is 6 sq units.

Page No 84:

Question 18:

Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5) such that $PR = \frac{3}{5} PQ$ .

Answer:

Given: $PR = \frac{3}{5} PQ$
$\Rightarrow \frac{PQ}{PR} = \frac{5}{3} \Rightarrow \frac{PR + RQ}{PR} = \frac{5}{3} \Rightarrow 1 + \frac{RQ}{PR} = \frac{5}{3} \Rightarrow \frac{RQ}{PR} = \frac{2}{3} \Rightarrow \frac{PR}{RQ} = \frac{3}{2}$
Let R(x, y) divide line segment PQ in the ratio 3 : 2.
Therefore, by section formula
$(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$
∴ (x, y) = $(\frac{3 (2) + 2 (- 1)}{3 + 2}, \frac{3 (5) + 2 (3)}{3 + 2})$
= $(\frac{4}{5}, \frac{21}{5})$

Hence, the coordinates of R is $(\frac{4}{5}, \frac{21}{5})$ .

Page No 84:

Question 19:

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Answer:

Given: A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear
Now, for three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) to be collinear
$△ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$
$∴ \frac{1}{2} [(k + 1) (2 k + 3 - 5 k) + 3 k (5 k - 2 k) + (5 k - 1) (2 k - 2 k - 3)] = 0 \Rightarrow - 3 k^{2} + 3 + 9 k^{2} - 15 k + 3 = 0 \Rightarrow 6 k^{2} - 15 k + 6 = 0 \Rightarrow 2 k^{2} - 5 k + 2 = 0 \Rightarrow (k - 2) (2 k - 1) = 0 \Rightarrow k = 2, \frac{1}{2}$

Hence, the required values of k are 2, $\frac{1}{2}$ .

Page No 84:

Question 20:

Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.

Answer:

Given: Points A(8, –9) and B(2, 1) and equation of line is 2x + 3y – 5 = 0
Let the point of division in the form be k : 1.
By section formula,
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n}$
∴ (x, y) = $(\frac{k \times 2 + 1 \times 8}{k + 1}, \frac{k \times 1 - 1 \times 9}{k + 1})$
= $(\frac{2 k + 8}{k + 1}, \frac{k - 9}{k + 1})$

Now, the above point lies on the line 2x + 3y – 5 = 0.
$2 (\frac{2 k + 8}{k + 1}) + 3 (\frac{k - 9}{k + 1}) - 5 = 0 \Rightarrow 2 (2 k + 8) + 3 (k - 9) = 5 (k + 1) \Rightarrow 2 k - 16 = 0 \Rightarrow k = 8$
$\begin{array}{rcl} ∴ (x, y) & = & (\frac{2 \times 8 + 8}{8 + 1}, \frac{8 - 9}{8 + 1}) \\ = & (\frac{24}{9}, \frac{- 1}{9}) \\ = & (\frac{8}{3}, \frac{- 1}{9}) \end{array}$

Hence, the coordinates of the point of division is $\begin{array}{rcl} (\frac{8}{3}, \frac{- 1}{9}) \end{array}$ .

Page No 85:

Question 1:

If (–4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answer:

Given: A(–4, 3) and B(4, 3)
Let the third vertex of the equilateral triangle be C(x, y).
AC = BC (triangle is equilateral)
$\sqrt{{(x + 4)}^{2} + {(y - 3)}^{2}} = \sqrt{{(x - 4)}^{2} + {(y - 3)}^{2}}$
Squaring both sides, we get
${(x + 4)}^{2} + {(y - 3)}^{2} = {(x - 4)}^{2} + {(y - 3)}^{2} \Rightarrow x^{2} + 16 + 8 x = x^{2} + 16 - 8 x \Rightarrow x = 0$
∴ The point is C(0, y).
By distance formula
$\begin{array}{rcl} AB & = & \sqrt{{(4 + 4)}^{2} + {(3 - 3)}^{2}} \\ = & \sqrt{8^{2}} \\ = & 8 \end{array}$

Now, AC = 8
$\Rightarrow \sqrt{{(4)}^{2} + {(y - 3)}^{2}} = 8 \Rightarrow y^{2} - 6 y - 39 = 0 y = \frac{6 \pm \sqrt{192}}{2} y = 3 - 4 \sqrt{3} (∵ Origin lies interior)$

Hence, the coordinates of C is $(0, 3 - 4 \sqrt{3})$ .

Page No 85:

Question 2:

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of âˆ†ADE.

Answer:

Given: A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD
Let the fourth vertex of parallelogram ABCD be (x, y)

Since, the diagonals of parallelogram bisect each other,
∴ Mid-point of BD = Mid-point of AC
$\Rightarrow (\frac{8 + x}{2}, \frac{2 + y}{2}) = (\frac{6 + 9}{2}, \frac{1 + 4}{2}) \Rightarrow (\frac{8 + x}{2}, \frac{2 + y}{2}) = (\frac{15}{2}, \frac{5}{2}) ∴ \frac{8 + x}{2} = \frac{15}{2} \Rightarrow 8 + x = 15 \Rightarrow x = 7 and \frac{2 + y}{2} = \frac{5}{2} \Rightarrow y = 3$
So, fourth vertex of parallelogram is D(7, 3).

Now, mid-point of DC = $(\frac{9 + 7}{2}, \frac{4 + 3}{2})$
E = $(8, \frac{7}{2})$
∴ Area of âˆ†ADE with vertices A(6, 1), D(7, 3) and E $(8, \frac{7}{2})$
$∆ = \frac{1}{2} [6 (3 - \frac{7}{2}) + 7 (\frac{7}{2} - 1) + 8 (1 - 3)] ∆ = \frac{1}{2} [6 \times (\frac{- 1}{2}) + 7 \times (\frac{5}{2}) + 8 \times (- 2)] ∆ = \frac{1}{2} [- 3 + (\frac{35}{2}) - 16] ∆ = \frac{- 3}{4}$
But, area can't be negative.

Hence, area of âˆ†ADE is $\frac{3}{4}$ sq units.

Page No 85:

Question 3:

The points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of âˆ†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 coordinates of points 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 centroid of the triangle ABC?

Answer:

Given : Coordinates of A, B and C of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃)

(i) We have, D is the mid-point of BC.

∴ D = $(\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2})$
(ii) Let the coordinates of point P be (x, y).
Since, point P divides line AD in the ratio 2 : 1
Then, by using internal section formula, we have $(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$ .

∴ Coordinates of P = $[\frac{2 \times (\frac{x_{2} + x_{3}}{2}) + 1 \times x_{1}}{2 + 1}, \frac{2 \times (\frac{y_{2} + y_{3}}{2}) + 1 \times y_{1}}{2 + 1}] = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

(iii) Let the coordinates of Q be (u, v).
Now, the line joining B(x₂, y₂) and E $(\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2})$ in the ratio (2 : 1)
Then, by using internal section formula, we have
Coordinates of Q = $[\frac{2 \times (\frac{x_{1} + x_{3}}{2}) + 1 \times x_{2}}{2 + 1}, \frac{2 \times (\frac{y_{1} + y_{3}}{2}) + 1 \times y_{2}}{2 + 1}] = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$
Also, let the coordinates of R(p, q)
Now, the line joining C(x₃, y₃) and F $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ in the ratio (2 : 1)
Then, by using internal section formula, we have
coordinates of R = $[\frac{2 \times (\frac{x_{1} + x_{2}}{2}) + 1 \times x_{3}}{2 + 1}, \frac{2 \times (\frac{y_{1} + y_{2}}{2}) + 1 \times y_{3}}{2 + 1}] = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

(iv) Coordinate of the centroid of âˆ†ABC = $(\frac{Sum of abscissae of ∆ ABC}{3}, \frac{Sum of ordinate of ∆ ABC}{3})$
= $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

Page No 85:

Question 4:

If the points A(1, –2), B(2, 3) C(a, 2) and D(–4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Answer:

We know, diagonals of a parallelogram bisect each other.
Thus, coordinates of midpoint of AC = coordinates of midpoint of BD
$\Rightarrow (\frac{a + 1}{2}, \frac{2 - 2}{2}) = (\frac{- 4 + 2}{2}, \frac{- 3 + 3}{2}) \Rightarrow (\frac{a + 1}{2}, 0) = (- 1, 0) \Rightarrow \frac{a + 1}{2} = - 1 \Rightarrow a + 1 = - 2 \Rightarrow a = - 3$

$\begin{array}{rcl} ∴ AB & = & \sqrt{{(2 - 1)}^{2} + {(3 + 2)}^{2}} \\ = & \sqrt{1^{2} + 5^{2}} \\ = & \sqrt{26} \end{array}$
Area of âˆ†ABC
$∆ = \frac{1}{2} [1 (3 - 2) + 2 (2 + 2) - 3 (- 2 - 3)] ∆ = \frac{1}{2} (1 + 8 + 15) ∆ = 12 sq units$
Thus, Area of Parallelogram = 2 × Area of âˆ†ABC
= 2 × 12
= 24 sq units
Also, Area of Parallelogram = Base × Height
⇒ 24 = $\sqrt{26}$ × Height
⇒ Height = $\frac{24}{\sqrt{26}} = \frac{12 \sqrt{26}}{13}$

Page No 85:

Question 5:

Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?

Answer:

Yes, here A, B, C and D are the vertices of a quadrilateral. Now, we will first find the type of quadrilateral

$AB= \sqrt{{(7 - 3)}^{2} + {(9 - 5)}^{2}} = \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2}$
$BC= \sqrt{{(11 - 7)}^{2} + {(5 - 9)}^{2}} = \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2} CD = \sqrt{{(7 - 11)}^{2} + {(1 - 5)}^{2}} = \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2} DA = \sqrt{{(3 - 7)}^{2} + {(5 - 1)}^{2}} = \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2}$
Also, the length of both diagonals
$AC = \sqrt{{(11 - 3)}^{2} + {(5 - 5)}^{2}} = \sqrt{8^{2} + 0} = 8 BD = \sqrt{{(7 - 7)}^{2} + {(1 - 9)}^{2}} = \sqrt{0 + 8^{2}} = 8$
Since AB = BC = CD = DA and AD = BC
Therefore, ABCD is a square. Also, diagonals of a square bisect each other.
So, P is a position of Jaspal in which he is equidistant from A, B, C and D.
∴ Coordinates of point P = Coordinates of the midpoint of AC
$= (\frac{3 + 11}{2}, \frac{5 + 5}{2}) = (7, 5)$

Hence, the required position of Jaspal is (7, 5).

Page No 85:

Question 6:

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 his 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 km.

Answer:

By distance formula, the distance between two points (x_1,y₁) and (x_2,y₂) is given by $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ .

Now, the distance between house and bank
$= \sqrt{{(5 - 2)}^{2} + {(8 - 4)}^{2}} = \sqrt{3^{2} + 4^{2}} = 5$
Distance between bank and daughter's school
$= \sqrt{{(13 - 5)}^{2} + {(14 - 8)}^{2}} = \sqrt{8^{2} + 6^{2}} = 10$
Distance between daughter's school to office
$= \sqrt{{(13 - 13)}^{2} + {(26 - 14)}^{2}} = \sqrt{0 + 12^{2}} = 12$
Thus, total distance travelled = 5 + 10 + 12
= 27 units
Now, distance between home and house
$= \sqrt{{(13 - 2)}^{2} + {(26 - 4)}^{2}} = \sqrt{11^{2} + 22^{2}} = \sqrt{605} \approx 24.6 units$
Hence, extra distance travelled by Ayush = 27 − 24.6
= 2.4 units

View NCERT Solutions for all chapters of Class 10