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(B) The equation of the line is $x-y=2$,which can be written as $\frac{x-y}{2}=1$. Substituting this into the homogeneous equation of the curve $5x^2+

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$xy=0$

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(B) The equation of the line is $x-y=2$,which can be written as $\frac{x-y}{2}=1$. Substituting this into the homogeneous equation of the curve $5x^2+12xy-8y^2+(8x-4y)(1) + 12(1)^2=0$: $5x^2+12xy-8y^2+(8x-4y)(\frac{x-y}{2}) + 12(\frac{x-y}{2})^2=0$. Multiplying by $4$ to clear the denominator: $20x^2+48xy-32y^2+2(8x-4y)(x-y)+3(x^2-2xy+y^2)=0$. $20x^2+48xy-32y^2+2(8x^2-12xy+4y^2)+3x^2-6xy+3y^2=0$. $20x^2+48xy-32y^2+16x^2-24xy+8y^2+3x^2-6xy+3y^2=0$. $39x^2+18xy-21y^2=0$. Dividing by $3$: $13x^2+6xy-7y^2=0$. Factoring: $(13x-7y)(x+y)=0$. The lines are $13x-7y=0$ and $x+y=0$. The angle bisectors of these lines are equally inclined to the coordinate axes,which corresponds to the pair of lines $xy=0$.

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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