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(A) The equation of the line is $x-y=2$,which can be written as $\frac{x-y}{2}=1$. Making the curve $5x^2+12xy-8y^2+8x-4y+12=0$ homogeneous using the

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$x^2-xy=0$

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(A) The equation of the line is $x-y=2$,which can be written as $\frac{x-y}{2}=1$. Making the curve $5x^2+12xy-8y^2+8x-4y+12=0$ homogeneous using the line equation: $5x^2+12xy-8y^2+(8x-4y)(\frac{x-y}{2})+12(\frac{x-y}{2})^2=0$. Expanding this: $5x^2+12xy-8y^2+(4x-2y)(x-y)+3(x^2-2xy+y^2)=0$. $5x^2+12xy-8y^2+4x^2-6xy+2y^2+3x^2-6xy+3y^2=0$. $12x^2-3y^2=0$,which simplifies to $4x^2-y^2=0$. The pair of lines is $y^2-4x^2=0$. The angle bisectors of $y^2-4x^2=0$ are given by $\frac{x^2-y^2}{1-(-4)} = \frac{xy}{0}$,which implies $x^2-y^2=0$. Comparing the given options,the pair of lines $x^2-xy=0$ (i.e.,$x(x-y)=0$) represents lines $x=0$ and $x=y$,which are the bisectors of the pair $xy=0$.

The pair of lines joining the origin to the points of intersection of the line $x-y=2$ with the curve $5x^2+12xy-8y^2+8x-4y+12=0$ are equally inclined to the pair of lines

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