If the lines ax^2 + 2hxy + by^2 = 0 represent | vedclass

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(B) Let the equations of the lines represented by $ax^2 + 2hxy + by^2 = 0$ be $y - m_1x = 0$ and $y - m_2x = 0$. One diagonal $AC$ is given by $lx + m

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$(am - hl)x = (bl - hm)y$

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(B) Let the equations of the lines represented by $ax^2 + 2hxy + by^2 = 0$ be $y - m_1x = 0$ and $y - m_2x = 0$. One diagonal $AC$ is given by $lx + my = 1$.From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.Solving the equations of $OA$ $(y = m_1x)$ and $OC$ $(y = m_2x)$ with the line $AC$ $(lx + my = 1)$,we get the coordinates of $A$ and $C$:$A = \left(\frac{1}{l + mm_1}, \frac{m_1}{l + mm_1}\right)$ and $C = \left(\frac{1}{l + mm_2}, \frac{m_2}{l + mm_2}\right)$.The midpoint $M$ of $AC$ is $\left(\frac{1}{2}\left(\frac{1}{l + mm_1} + \frac{1}{l + mm_2}\right), \frac{1}{2}\left(\frac{m_1}{l + mm_1} + \frac{m_2}{l + mm_2}\right)\right)$.The second diagonal passes through the origin $O(0,0)$ and the midpoint $M$. The slope of this diagonal is $\frac{y_M}{x_M} = \frac{m_1(l + mm_2) + m_2(l + mm_1)}{(l + mm_2) + (l + mm_1)} = \frac{l(m_1 + m_2) + 2mm_1m_2}{2l + m(m_1 + m_2)}$.Substituting $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$,the slope is $\frac{l(-2h/b) + 2m(a/b)}{2l + m(-2h/b)} = \frac{-2hl + 2am}{2bl - 2hm} = \frac{am - hl}{bl - hm}$.Thus,the equation of the diagonal is $y = \left(\frac{am - hl}{bl - hm}\right)x$,which simplifies to $(bl - hm)y = (am - hl)x$.

If the lines $ax^2 + 2hxy + by^2 = 0$ represent the adjacent sides of a parallelogram,then the equation of the second diagonal,if one diagonal is $lx + my = 1$,will be

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