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(C) The given equations are $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$.For the first equation: $x(y+4)-3(y+4)=0 \Rightarrow (x-3)(y+4)=0$. This represents th

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

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(C) The given equations are $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$.For the first equation: $x(y+4)-3(y+4)=0 \Rightarrow (x-3)(y+4)=0$. This represents the lines $x=3$ and $y=-4$.For the second equation: $x(y-3)+4(y-3)=0 \Rightarrow (x+4)(y-3)=0$. This represents the lines $x=-4$ and $y=3$.The four lines forming the square are $x=3, x=-4, y=3, y=-4$.The vertices of the square are $A(-4,-4), B(3,-4), C(3,3),$ and $D(-4,3)$.The diagonal $AC$ passes through $A(-4,-4)$ and $C(3,3)$. Its equation is $y-(-4) = \frac{3-(-4)}{3-(-4)}(x-(-4))$ $\Rightarrow y+4 = 1(x+4)$ $\Rightarrow x-y=0$.The diagonal $BD$ passes through $B(3,-4)$ and $D(-4,3)$. Its equation is $y-(-4) = \frac{3-(-4)}{-4-3}(x-3)$ $\Rightarrow y+4 = \frac{7}{-7}(x-3)$ $\Rightarrow y+4 = -x+3$ $\Rightarrow x+y+1=0$.The combined equation of the diagonals is $(x-y)(x+y+1)=0$.Expanding this,we get $x^2+xy+x-xy-y^2-y=0$,which simplifies to $x^2-y^2+x-y=0$.

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

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