Question

The equation of the diagonal of the square formed by the pairs of lines $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ is

• A. $x-y=0$
• B. $x+y+1=0$
• C. $x+y=0$
• D. $x-y+1=0$

Answer 1

The correct option is B $x+y+1=0$

Given, first equation of pair of lines

$\Rightarrow xy+4x-3y-12=0$

$\Rightarrow x(y+4)-3(y+4)=0$

$\Rightarrow (x-3)(y+4)=0$

Hence, this is a pair of lines of $x-3=0,y+4=0$

So point of intersection of these two lines is $P_1(3,-4)$

Considering second pair of lines

$\Rightarrow xy-3x+4y-12=0$

$\Rightarrow x(y-3)+4(y-3)=0$

$\Rightarrow (x+4)(y-3)=0$

Hence, this is a pair of lines of $x+4=0,y-3=0$

So point of intersection of these two lines is $P_2(-4,3)$

Now applying two points formula for $P_{1}and{P}_2$ in order to get the equation of the diagonal

$\Rightarrow x-x_1=\frac{y_2-y_1}{x_2-x_1}(y-y_1)$

$\Rightarrow x-3=\frac{3-(-4)}{(-4)-3}(y-(-4\left)\right)$

$\Rightarrow x-3=\frac{7}{(-7)}(y+4)$

$\Rightarrow x-3=-y-4$

$\Rightarrow x+y+1=0$

Hence, answer is option $B$