Question: A straight line through origin bisect the line passing through the given points $ (a\cos \alpha ,a\sin \alpha ) $ and $ (a\cos \beta ,a\sin \beta ) $ , then the lines are
Options:
A) Perpendicular
B) Parallel
C) Angle between them is $ \frac{\pi }{4} $
D) None of these
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Answer:
Correct Answer: A
Solution:
- Mid point of $ (a\cos \alpha ,a\sin \alpha ) $ and $ (a\cos \beta ,a\sin \beta ) $ is $ P( \frac{a(\cos \alpha +\cos \beta )}{2},\frac{a(\sin \alpha +\sin \beta )}{2} ) $ Slope of line $ AB $ is $ \frac{a\sin \beta -a\sin \alpha }{a\cos \beta -a\cos \alpha } $ $ =\frac{\sin \beta -\sin \alpha }{\cos \beta -\cos \alpha }=m_1 $ and slope of $ OP $ is $ \frac{\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }=m_2 $ Now $ m_1\times m_2=\frac{{{\sin }^{2}}\beta -{{\sin }^{2}}\alpha }{{{\cos }^{2}}\beta -{{\cos }^{2}}\alpha }=-1 $ Hence the lines are perpendicular.
BETA
