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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
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Perpendicular
Mid point of $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ is $P\left(\frac{a(\cos \alpha+\cos \beta)}{2}, \frac{a(\sin \alpha+\sin \beta)}{2}\right)$
Slope of line $A B$ 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 $O P$ 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.
Slope of line $A B$ 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 $O P$ 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.
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