If 2x^2+3xy-2y^2=0 represents two sides of a | vedclass

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(C) The given pair of lines representing two sides of a parallelogram is $2x^2+3xy-2y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we have $a=2, b=-2,

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$x-3y=0$

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(C) The given pair of lines representing two sides of a parallelogram is $2x^2+3xy-2y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we have $a=2, b=-2, h=\frac{3}{2}$. The given diagonal is $3x+y=-1$,which can be written as $lx+my=1$ where $l=3, m=1$ (by dividing by $-1$). The formula for the other diagonal of a parallelogram formed by $ax^2+2hxy+by^2=0$ and diagonal $lx+my=1$ is $y(bl-hm) = x(am-hl)$. Substituting the values: $y((-2)(3) - (\frac{3}{2})(1)) = x((2)(1) - (\frac{3}{2})(3))$ $y(-6 - \frac{3}{2}) = x(2 - \frac{9}{2})$ $y(-\frac{15}{2}) = x(-\frac{5}{2})$ $15y = 5x$ $3y = x \Rightarrow x-3y=0$.

If $2x^2+3xy-2y^2=0$ represents two sides of a parallelogram and $3x+y+1=0$ is one of its diagonals,then the other diagonal is

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