the x coordinate of point P is twice its y coordinate . if P is equidistant from Q(2,-5) R(-3,6).then find the coordinates of point P

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Solution

$Let P (x, y) is the required point . Let Q (2, - 5) and R (- 3, 6) are the given points . Now, PQ = PR \Rightarrow {PQ}^{2} = {PR}^{2} \Rightarrow {(x - 2)}^{2} + {(y + 5)}^{2} = {(x + 3)}^{2} + {(y - 6)}^{2} [Using distance formula] \Rightarrow x^{2} + 4 - 4 x + y^{2} + 25 + 10 y = x^{2} + 9 + 6 x + y^{2} + 36 - 12 y \Rightarrow - 4 x + 10 y + 29 = 6 x - 12 y + 45 \Rightarrow - 10 x + 22 y - 16 = 0 \Rightarrow - 10 (2 y) + 22 y = 16 [a s, x = 2 y] \Rightarrow - 20 y + 22 y = 16 \Rightarrow 2 y = 16 \Rightarrow y = 8 Now, x = 2 y = 2 \times 8 = 16 So, coordinates of required point P are P (16, 8)$

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