The equation of the pair of straight lines through the point $(1, 1)$ and perpendicular to the pair of lines and perpendicular to the pair of lines $3 x^{2} - 8 x y + 5 y^{2} = 0$ is

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$5 x^{2} + 8 x y + 3 y^{2} - 14 x - 18 y + 16 = 0$

$5 x^{2} - 8 x y + 3 y^{2} - 18 x - 14 y + 32 = 0$

$5 x^{2} + 8 x y + 3 y^{2} - 18 x - 14 y + 16 = 0$

$5 x^{2} - 8 x y + 3 y^{2} - 14 x - 18 y + 32 = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 3 x^{2} - 8 x y + 5 y^{2} = 0 - - (1) \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} = 0 \\ a = 3, h = - 4, b = 5 \\ E q u a t i o n o f t h e l i n e p a s \sin g t h r o u g h (1, 1) a n d ⊥ r t o (1) i s \\ b {(x - x_{1})}^{2} - 2 h (x - x_{1}) (y - y_{1}) + {(y - y_{1})}^{2} = 0 \\ \Rightarrow 5 {(x - 1)}^{2} + 8 (x - 1) (y - 1) + 3 {(y - 1)}^{2} = 0 \\ \Rightarrow 5 (x^{2} + 1 - 2 x) + 8 (x y - x - y + 1) + 3 (y^{2} + 1 - 2 y) = 0 \\ \Rightarrow 5 x^{2} + 5 - 10 x + 8 x y - 8 x - 8 y + 8 + 3 y^{2} + 3 - 6 y = 0 \\ \Rightarrow 5 x^{2} + 8 x y + 3 y^{2} - 18 x - 14 y + 16 = 0 \end{array}$