Pair Of Straight Lines Question 104
Question: If $ y=mx $ be one of the bisectors of the angle between the lines $ ax^{2}-2hxy+by^{2}=0 $ , then
Options:
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 $
Show Answer
Answer:
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 -mx^{2}-xy(m^{2}-1)+my^{2}=0 $ …(i)
Also equations of bisectors of $ ax^{2}-2hxy+by^{2}=0 $ is $ -hx^{2}-(a-b)xy+hy^{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 $ .
BETA
