The combined equation of the diagonals of the square formed by the pairs of lines xy + 6y – 4x - 24 = 0 and
xy + 6x - 4y -24 = 0

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

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

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

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

answer is A.

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Detailed Solution

Given Equation

$\begin{array}{l} x y + 6 y - 4 x - 24 = 0 \\ \Rightarrow y (x + 6) - 4 (x + 6) = 0 \\ \Rightarrow (x + 6) (y - 4) = 0 \\ \Rightarrow x + 6 = 0 - - (1), y - 4 = 0 - - (2) \\ a n d \\ x y + 6 x - 4 y - 24 = 0 \\ \Rightarrow x (y + 6) - 4 (y + 6) = 0 \\ \Rightarrow (x - 4) (y + 6) = 0 \\ \Rightarrow x - 4 = 0 - - (3) \end{array}$

A (-6, 4) B(4, 4) C(4 , -6) D(-6 ,-6)

$\begin{array}{l} E q u a t i o n o f \bar{A C} \Rightarrow y - 4 = \frac{- 6 - 4}{4 + 6} (x + 6) \\ \Rightarrow y - 4 = - x - 6 \\ \Rightarrow x + y + 2 = 0 \\ E q u a t i o n o f \bar{B D} \Rightarrow y - 4 = \frac{- 6 - 4}{- 6 - 4} (x - 4) \\ \Rightarrow y - 4 = x - 4 \\ \Rightarrow x - y = 0 \\ C o m b a i n e d E q u a t i o n i s (x - y) (x + y + 2) = 0 \\ \Rightarrow x^{2} - y^{2} + 2 x - 2 y = 0 \end{array}$