$\begin{array}{l} T h e e q u a t i o n t o t h e p a i r o f s t r a i g h t l i n e s p a s \sin g t h r o u g h (2, 1) a n d p e r p e n d i c u l a r t o \\ t h e p a i r o f l i n e s 4 x y + 2 x + 6 y + 3 = 0 i s \end{array}$

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$x y + x + 2 y + 2 = 0$

$x y + x + 2 y - 2 = 0$

$x y + x - 2 y - 2 = 0$

$x y - x - 2 y + 2 = 0$

answer is D.

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Detailed Solution

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