A) \[h\,(1+{{m}^{2}})+m(a-b)=0\]
B) \[h\,(1-{{m}^{2}})+m(a+b)=0\]
C) \[h\,(1-{{m}^{2}})+m(a-b)=0\]
D) \[h\,(1+{{m}^{2}})+m(a+b)=0\]
Correct Answer: C
Solution :
Here equation of one bisector of angle is \[y-mx=0,\] therefore equation of second is \[x+my=0\]. Hence combined equation is \[(x+my)(y-mx)=0\] \[\Rightarrow -m{{x}^{2}}-xy({{m}^{2}}-1)+m{{y}^{2}}=0\] .?.(i) Also equations of bisectors of \[a{{x}^{2}}-2hxy+b{{y}^{2}}=0\] is \[-h{{x}^{2}}-(a-b)xy+h{{y}^{2}}=0\] .....(ii) Hence (i) and (ii) are the same equations, therefore \[\frac{m}{h}=\frac{{{m}^{2}}-1}{(a-b)}\Rightarrow h({{m}^{2}}-1)=m(a-b)\] \[\Rightarrow m(a-b)+h(1-{{m}^{2}})=0\].
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