solve Q 14 plz dont give link

solve Q 14 plz dont give link 14. 15. 16. 17. 18. Consider the points P(2,—4); Q(4,—2) and R(7, I The points Y (A) form an equilateral triangle (B) form a right angled triangle (C) form an isosceles triangle which is not equilateral (D) are collinear. A triangle has two of its vertices at (0,1) and (2,2) in the cartesian plane. Its third vertex lies on the x-axis. Ifthe area of the triangle is 2 square units then the sum of the possible abscissae of the third vertex, is- A point P(x,y) moves so that the sum of the distance from P to the coordinate axes is equal to the distance from P to the point A(I, I). The equation of the locus of P in the first quadrant is Let A(2,—3) and B(—2,1) be vertices of a AABC. If the centroid of AABC moves on the line 2x + 3y —l, then the locus of the vertex C is- (B) 2x -3y=7 (A) 2x + 3y = 9 (D) 3x - 2y=3 A stick of length 10 units rests against the floor and a wall ofa room. Ifthe stick begins to slide on the floor then the locus of its middle point is : (A) x2 + = 2.5 (B) x2 + Y2 = 25 (c) + = 100 (D) none Given the points A(0,4) and B(0,—4), the equation of the locus of the point P such that IAP—

Dear Student,

L e t t h e v e r t e x o f t r i a n g l e b e A (0, 1), B (2, 2) a n d C (a, b) \sin c e v e r t e x C l i e s o n x - a x i s . h e n c e y = 0 t h e r e f o r e c o o r d i n a t e s o f C (a, 0) N o w a r e a o f t r i a n g l e = \frac{1}{2} \times |\begin{matrix} 0 & 1 & 1 \\ 2 & 2 & 1 \\ a & 0 & 1 \end{matrix}| \Rightarrow \frac{1}{2} \times [0 \times (2 \times 1 - 1 \times 0) - 1 \times (2 \times 1 - a \times 1) + 1 \times (2 \times 0 - 2 \times a)] \Rightarrow \frac{1}{2} \times [0 \times (2) - 1 \times (2 - a) + 1 \times (- 2 \times a)] \Rightarrow \frac{1}{2} \times [0 - 1 \times (2 - a) + 1 \times (- 2 \times a)] \Rightarrow \frac{1}{2} \times [a - 2 - 2 a] \Rightarrow \frac{1}{2} \times [- 2 - a] \Rightarrow \frac{1}{2} \times |- 1 \times (2 + a)| \Rightarrow \frac{1}{2} (2 + a) = 2 b e c a u s e a r e a o f t r i a n g l e i s g i v e n a s 2 2 + a = 4 a = 2 - - (1) b e c a u s e t h e m o d c a n a c c e p t t w o v a l u e s (2 + a) a n d (- 2 - a) . H e n c e \frac{1}{2} (- 2 - a) = 2 - 2 - a = 4 - a = 6 a = - 6 - - (2) A d d i n g p o s s i b l e v a l u e s o f a f r o m e q u a t i o n 1 a n d 2 2 - 6 = - 4 H e n c e t h e s u m o f a b s c i s s a e (p o s s i b l e v a l u e s o f a) = - 4

Option A is correct

Regards,