If lx^2+3xy-2y^2-5x+5y+k=0 represents a pair | vedclass

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(C) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. For the equation $lx^2+3xy-2y^2-5x+5y+k=0$ to represent a pair of p

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$k=-3, l=2$

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(C) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. For the equation $lx^2+3xy-2y^2-5x+5y+k=0$ to represent a pair of perpendicular lines,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$. Here,$a=l$ and $b=-2$. So,$l+(-2)=0 \Rightarrow l=2$. Now the equation becomes $2x^2+3xy-2y^2-5x+5y+k=0$. For this to represent a pair of lines,the determinant condition $abc+2fgh-af^2-bg^2-ch^2=0$ must be satisfied. Here,$a=2, b=-2, c=k, h=\frac{3}{2}, g=-\frac{5}{2}, f=\frac{5}{2}$. Substituting these values: $(2)(-2)(k) + 2(\frac{5}{2})(-\frac{5}{2})(\frac{3}{2}) - 2(\frac{5}{2})^2 - (-2)(-\frac{5}{2})^2 - k(\frac{3}{2})^2 = 0$. $-4k - \frac{75}{4} - \frac{50}{4} + \frac{50}{4} - \frac{9k}{4} = 0$. $-4k - \frac{9k}{4} - \frac{75}{4} = 0$. $-\frac{25k}{4} = \frac{75}{4} \Rightarrow k=-3$. Thus,$l=2$ and $k=-3$.

If $lx^2+3xy-2y^2-5x+5y+k=0$ represents a pair of perpendicular lines,then

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