The pair of lines joining the origin to the points of intersection of the line x – y = 2 with the curve $5 x^{2} + 12 x y - 8 y^{2} - 4 y + 12 = 0$ are equally inclined to the pair of lines

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

$(x - 2) (y - 2) = 0$

$x y - y^{2} = 4$

$x y = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 5 x^{2} + 12 x y - 8 y^{2} + 8 x - 4 y + 12 = 0 - - (1) \\ a n d t h e l i n e x - y = 2 \\ \Rightarrow \frac{x - y}{2} = 1 - - - (2) \\ H o m o g e n i z i n g (1) w i t h (2) \\ 5 x^{2} + 12 x y - 8 y^{2} + 8 x (\frac{x - y}{2}) - 4 y (\frac{x - y}{2}) + 12 {(\frac{x - y}{2})}^{2} = 0 \\ \Rightarrow 5 x^{2} + 12 x y - 8 y^{2} + 4 x (x - y) - 2 y (x - y) + 12 (\frac{x^{2} + y^{2} - 2 x y}{4}) = 0 \\ \Rightarrow 5 x^{2} + 12 x y - 8 y^{2} + 4 x^{2} - 4 x y - 2 x y + 2 y^{2} + 3 x^{2} + 3 y^{2} - 6 x y = 0 \\ \Rightarrow 12 x^{2} - 3 y^{2} = 0 \\ \Rightarrow 4 x^{2} - y^{2} = 0 \\ T h e l i n e s E q u a l l y i n c l i n e d w i t h x y = 0 \end{array}$