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(D) The first pair of lines is $xy+4x-3y-12=0$,which factors as $(x-3)(y+4)=0$. This represents the lines $x=3$ and $y=-4$.The second pair of lines is

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The first pair of lines is $xy+4x-3y-12=0$,which factors as $(x-3)(y+4)=0$. This represents the lines $x=3$ and $y=-4$.The second pair of lines is $xy-3x+4y-12=0$,which factors as $(x+4)(y-3)=0$. This represents the lines $x=-4$ and $y=3$.The four lines forming the square are $x=3, x=-4, y=-4, y=3$.The vertices of the square are $(3,3), (3,-4), (-4,-4), (-4,3)$.The diagonals connect $(3,3)$ to $(-4,-4)$ and $(3,-4)$ to $(-4,3)$.The equation of the diagonal passing through $(3,3)$ and $(-4,-4)$ is $y-3 = \frac{-4-3}{-4-3}(x-3)$,which simplifies to $y-3 = x-3$,or $x-y=0$.The equation of the diagonal passing through $(3,-4)$ and $(-4,3)$ is $y-(-4) = \frac{3-(-4)}{-4-3}(x-3)$,which simplifies to $y+4 = -1(x-3)$,or $x+y+1=0$.The combined equation is $(x-y)(x+y+1) = 0$,which is $x^2-y^2+x-y=0$.

The combined equation of the diagonals of the square formed by the two pairs of straight lines given by $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ is

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