• # question_answer If the lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents the adjacent sides of a parallelogram, then the equation of second diagonal if one is $lx+my=1$, will be  A)            $(am+hl)x=(bl+hm)y$    B)            $(am-hl)x=(bl-hm)y$ C)            $(am-hl)x=(bl+hm)y$     D)            None of these

Let the equation of lines represented by $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ be $y-{{m}_{1}}x=0$ and $y-{{m}_{2}}x=0$ and one diagonal AC  be $lx+my=1.$            Therefore ${{m}_{1}}+{{m}_{2}}=\frac{-2h}{b}$and ${{m}_{1}}{{m}_{2}}=\frac{a}{b}$ Now on solving the equation of OA and OC with the line AC, we get the coordinates of            $A\left( \frac{1}{l+m{{m}_{1}}},\frac{{{m}_{1}}}{l+m{{m}_{1}}} \right)$            and $C\left( \frac{1}{l+m{{m}_{2}}},\frac{{{m}_{2}}}{l+m{{m}_{2}}} \right)$            Now find the coordinates of mid-point of AC and the equation of diagonal through this mid-point and origin. The required equation is $x(am-hl)=(lb-mh)y$ .
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