The equation of the pair of lines through (1, -1) and perpendicular to the pair of lines $x^{2} - x y - 2 y^{2} = 0$ is

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$2 x^{2} - x y + y^{2} + 5 x + y + 2 = 0$

$2 x^{2} - x y - y^{2} + 5 x + y - 2 = 0$

$x^{2} - x y + 2 y^{2} - 5 x - y - 2 = 0$

$2 x^{2} - x y - y^{2} - 5 x - y + 2 = 0$

answer is B.

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Detailed Solution

Given Equation $x^{2} - x y - 2 y^{2} = 0$

Comparing with $a x^{2} + 2 h x y + b y^{2} = 0$

$a = 1, h = \frac{- 1}{2}, b = - 2$

Let The point $(1, - 1) = (x_{1}, y_{1})$

Required equation is $b {(x - x_{1})}^{2} - 2 h (x - x_{1}) (y - y_{1}) + a {(y - y_{1})}^{2} = 0$

$\begin{array}{l} \Rightarrow - 2 {(x - 1)}^{2} + (x - 1) (y + 1) + 1 {(y + 1)}^{2} = 0 \\ \Rightarrow 2 (x^{2} + 1 - 2 x) - (x y + x - y - 1) - (y^{2} + 1 + 2 y) = 0 \\ \Rightarrow 2 x^{2} + 2 - 4 x - x y - x + y + 1 - y^{2} - 1 - 2 y = 0 \\ \Rightarrow 2 x^{2} - x y - y^{2} - 5 x - y + 2 = 0 \end{array}$