Let PQR be a right angle isosceles triangle, right angle at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is

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$3 x^{2} - 3 y^{2} + 8 x y + 20 x - 10 y + 25 = 0$

$3 x^{2} - 3 y^{2} + 8 x y - 20 x + 10 y + 25 = 0$

$3 x^{2} - 3 y^{2} + 8 x y - 20 x - 10 y + 25 = 0$

$3 x^{2} - 3 y^{2} - 8 x y - 20 x - 10 y + 25 = 0$

answer is A.

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Detailed Solution

Given QR line 2x+y-3=0

$\begin{array}{l} m_{2} = s l o p e = \frac{- 2}{1} = - 2 \\ a n d P Q = P R a n d ∠ Q P R = 90^{0} \\ ∴ θ = 45^{0} \\ L e t m_{1} = m_{2} b e t h e s l o p e \bar{P Q} o r P R \\ ∴ Tan 45^{0} = |\frac{m + 2}{1 + m (- 2)}| \\ \Rightarrow 1 = |\frac{m + 2}{1 - 2 m}| \\ \Rightarrow \frac{m + 2}{1 - 2 m} = \pm 1 \\ \Rightarrow m + 2 = 1 - 2 m m + 2 = - (1 - 2 m) \\ \Rightarrow 3 m = - 1 m + 2 = - 1 + 2 m \\ \Rightarrow m = \frac{- 1}{3} m = 3 \end{array}$

If m=-1/3 and (2, 1)

Equation of $\bar{P Q} i s y - 1 = \frac{- 1}{3} (x - 2)$

$\begin{array}{l} \Rightarrow 3 y - 3 = - x + 2 \\ \Rightarrow x + 3 y - 5 = 0 \\ I f m = 3 a n d p (2, 1) \\ E q u a t i o n o f P R i s y - 1 = 3 (x - 2) \\ \Rightarrow 3 x - y - 5 = 0 \\ C o m b i n e d E q o f \bar{P Q} a n d \bar{P R} i s (x + 3 y - 5) (3 x - y - 5) = 0 \\ \Rightarrow 3 x^{2} + 8 x y - 3 y^{2} - 20 x - 10 y + 25 = 0 \end{array}$