Question

Show that the perpendicular from the origin upon the straight line joining the points $(a\cos \alpha ,a\sin \alpha )$ and $(a\cos \beta ,a\sin \beta )$ bisects the distance between them.

Answer 1

Equation of line joining $(a\cos \alpha ,a\sin \alpha )$ and $(a\cos \beta ,a\sin \beta )$ is

$y-a\sin \alpha =\left(\frac{a\sin \beta -a\sin \alpha }{a\cos \beta -a\cos \alpha }\right)(x-a\cos \alpha )$

$y-a\sin \alpha =\left(\frac{\sin \beta -\sin \alpha }{\cos \beta -\cos \alpha }\right)(x-a\cos \alpha )$

$y-a\sin \alpha =\left(\frac{2\cos \frac{\beta +\alpha }{2}\sin \frac{\beta -\alpha }{2}}{-2\sin \frac{\beta +\alpha }{2}\sin \frac{\beta -\alpha }{2}}\right)(x-a\cos \alpha )$

$y-a\sin \alpha =-\left(\frac{\cos \frac{\beta +\alpha }{2}}{\sin \frac{\beta +\alpha }{2}}\right)(x-a\cos \alpha )$

$\sin \frac{\beta +\alpha }{2}y-a\sin \alpha \sin \frac{\beta +\alpha }{2}=-\cos \frac{\beta +\alpha }{2}x+a\cos \alpha \cos \frac{\beta +\alpha }{2}$

$x\cos \frac{\beta +\alpha }{2}+y\sin \frac{\beta +\alpha }{2}=a(\cos \alpha \cos \frac{\beta +\alpha }{2}+\sin \alpha \sin \frac{\beta +\alpha }{2})$

$x\cos \frac{\beta +\alpha }{2}+y\sin \frac{\beta +\alpha }{2}=a\left(\cos (\alpha -\frac{\beta +\alpha }{2})\right)$

$x\cos \frac{\beta +\alpha }{2}+y\sin \frac{\beta +\alpha }{2}=a\cos \frac{\alpha -\beta }{2}$

Let the length of perpendicular from origin be $p$

$p=\frac{|0\left(\cos \frac{\beta -\alpha }{2}\right)-0\left(\sin \frac{\beta -\alpha }{2}\right)-a\cos \frac{\alpha -\beta }{2}|}{\sqrt{{\cos }^2\frac{\beta +\alpha }{2}+{\sin }^2\frac{\beta +\alpha }{2}}}p=\frac{|-a\cos \frac{\alpha -\beta }{2}|}{1}\Rightarrow p=a\cos \frac{\alpha -\beta }{2}$