Question: Line $ ax+by+p=0 $ makes angle $ \pi /4 $ with $ xcos\alpha +ysin\alpha =p,p\in {R^{+}} $ . if these lines and the line $ xsin\alpha -ycos\alpha =0 $ are concurrent, then
Options:
A) $ a^{2}+b^{2}=1 $
B) $ a^{2}+b^{2}=2 $
C) $ 2(a^{2}+b^{2})=1 $
D) none of these
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Answer:
Correct Answer: B
Solution:
- [b] Lines $ x\cos a+y\sin a=p $ and $ x\sin \alpha -y\cos \alpha =0 $ are mutually perpendicular, thus , $ ax+by+p=0 $ will be equally inclined to these lines and would be the angle bisector of these lines. Now, the equations of angle bisectors are $ x\sin \alpha -y\cos \alpha =\pm (xcos\alpha +ysin\alpha -p) $ Or $ x(cos\alpha -sin\alpha )+y(sin\alpha +cos\alpha )=p $ Or $ x(sin\alpha +cos\alpha )-y(cos\alpha -sin\alpha )=p $ Comparing th4ese lines with $ ax+by+p=0 $ , we get $ \frac{a}{\cos \alpha -\sin \alpha }=\frac{b}{\sin \alpha +\cos \alpha }=1 $ Or $ a^{2}+b^{2}=2 $ Or $ \frac{a}{\sin \alpha +\cos \alpha }=\frac{b}{\cos \alpha -\sin \alpha }=1 $ Or $ a^{2}+b^{2}=2 $
BETA
