If two pairs of straight lines with combined equations $\quad x y+4 x-3 y-12=0$ and $x y-3 x+4 y-12=0$ form a square. Then, the combined equation of its diagonals is

• A

  $x^2-2 x y+y^2+x-y=0$
• B

  $x^2+2 x y+y^2+x+y=0$
• Correct answer
  C

  $x^2-y^2+x-y=0$
• D

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

Solution

Given equation $x y+4 x-3 y-12=0$ can be written as $$ \begin{aligned} x(y+4)-3(y+4) & =0 \\ \Rightarrow \quad(y+4)(x-3) & =0 \end{aligned} $$ Also, the equation $x y-3 x+4 y-12=0$ can be written as $$ \begin{aligned} x(y-3)+4(y-3) & =0 \\ \Rightarrow \quad(y-3)(x+4) & =0 \end{aligned} $$ Hence, equations of the sides of the square are $$ x=3, x=-4, y=3, y=-4 $$ So, the coordinates of the vertex are $A(-4,-4)$, $B(3,-4), C(3,3)$ and $D(-4,3)$.
$$ \begin{aligned} & \therefore \text { Equation of } A C: y+4=\frac{3+4}{3+4}(x+4) \\ & \Rightarrow \quad y+4=x+4 \Rightarrow y=x \\ & \Rightarrow \quad x-y=0 \\ & \therefore \quad \text { Equation } B D: y+4=\frac{3+4}{-4-3}(x-3) \\ & \Rightarrow \quad y+4=-(x-3) \\ & \Rightarrow \quad y+4=-x+3 \\ & \Rightarrow \quad x+y+1=0 \\ & \end{aligned} $$ Hence, combined equation of diagonals is $$ \begin{array}{rlrl} & & (x-y)(x+y+1) & =0 \\ \Rightarrow & x^2+x y+x-x y-y^2-y & =0 \\ \Rightarrow & & x^2-y^2+x-y & =0 \end{array} $$

Hence, the correct answer is option: 'C'.