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
Correct Answer: B
Solution :
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 \[x(am-hl)=(lb-mh)y\] .You need to login to perform this action.
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