If A and B are two points having coordinates (3 , 4) and (5 , -2) respectively and P is a point such that PA = PB and area of PAB = 10 sq. units, then coordinates of P are

(7, 4)

(13, 2)

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(7, 2)

(1, 0)

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(2, 7)

(4, 13)

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(7, 4)

(13, 4)

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Solution

The correct option is B $(7, 2)$ & $(1, 0)$
$\begin{matrix} G i v e n t h a t P A = P B a r e a Δ P A B = 10 s q . u n i t L e t p \equiv (x, y) ∴ P A = P B \Rightarrow P A^{2} = P B^{2} \Rightarrow {(x - 3)}^{2} + {(y - 4)}^{2} = {(x - 5)}^{2} + {(y + 2)}^{2} \Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = x^{2} - 10 x + 25 + y^{2} + 4 y + 4 \Rightarrow 4 x - 12 y = 4 ∴ x - 3 y = 1 - - - (1) a g a i n a r e a (Δ P A B) = 10 \Rightarrow \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 10 \Rightarrow | x (- 2 - 4) + 5 (4 - y) + 3 (y - (- 2)) | = 20 ∴ 6 x - 2 y + 26 = \pm 20 F r o m t h e q u e s t i o n w e g e t \Rightarrow - 6 x - 2 y + 26 = 20 & \Rightarrow - 6 x - 2 y + 26 = - \Rightarrow - 6 x - 2 y = - 6 & \Rightarrow - 6 x - 2 y = - 46 \Rightarrow 3 x + y = 3 - - (2) & \Rightarrow 3 x + y = 23 - - - (3) F r o m e q u^{n} (1) a n d (2) w e h a v e x - 3 t = 1 3 x + y = 3 ∴ x = 1 a n d y = 0 f r o m e q u^{n} (1) a n d (3), w e g e t x - 3 y = 1 3 x + y = 23 ∴ x = 7 a n d ∴ x = 2 ∴ R e q u i r e d c o - o r d i n a t e o f p o i n t P a r e (1, 0) & (7, 2) H e n c e, o p t i o n B i s c o r r e c t . \end{matrix}$

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