Question

Find perpendicular distance from the origin of the line joining the points $(\cos \theta ,\sin \theta )$ and $(\cos \varphi ,\sin \varphi )$.

Answer 1

Let us consider the problem

$\frac{x-\cos \theta }{\cos \varphi -\cos \theta }=\frac{y-\sin \theta }{\sin \varphi -\sin \theta }$

implies that, $(x-\cos \theta )(\sin \varphi -\sin \theta )=(y-\sin \theta )(\cos \varphi -\cos \theta )$

implies that, $x(\sin \varphi -\sin \theta )-y(\cos \varphi -\cos \theta )-\cos \theta \sin \varphi +\cos \theta \sin \theta +\sin \theta \cos \varphi -\sin \theta \cos \theta =0$

But $\cos \varphi \sin \theta -\cos \theta \sin \varphi =\sin (\varphi -\theta )$

Therefore, $x(\sin \varphi -\sin \theta )-y(\cos \varphi -\cos \theta )+\sin (\varphi -\theta )=0$

On comparing with the equations $Ax+By+C=0$

$A=\sin \theta -\sin \varphi ,B=\cos \varphi -\cos \theta andC=\sin (\varphi -\theta )$

So, perpendicular distance $d$ of the line $Ax+By+C=0$ from a point $(x_1,y_1)$ is $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

Substitutes the values we get,

$d=\frac{|(\sin \theta -\sin \varphi )\left(0\right)+(\cos \varphi -\cos \theta )\left(0\right)+\sin (\varphi -\theta )|}{\sqrt{{(\sin \theta -\sin \varphi )}^2+{(\cos \varphi -\cos \theta )}^2}}$

$=\frac{|\sin \theta -\varphi |}{\sqrt{{\sin }^2\theta +{\sin }^2\varphi -2\sin \theta \sin \varphi +{\cos }^2\theta +{\cos }^2\varphi -2\cos \theta \cos \varphi }}$

$=\frac{|\sin (\theta -\varphi )|}{\sqrt{({\sin }^2\theta +{\cos }^2\theta )+({\sin }^2\varphi +{\cos }^2\varphi )-2(\sin \theta \sin \varphi +\cos \theta \cos \varphi )}}$

$=\frac{|\sin (\varphi -\theta )|}{\sqrt{1+1-2\cos (\varphi -\theta )}}$

$=\frac{|\sin (\varphi -\theta )|}{\sqrt{2(1-\cos (\varphi -\theta )}}$

The hint given is, $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$

Hence the required result is, $=\frac{|\sin (\varphi -\theta )|}{|2\sin \left(\frac{\varphi -\theta }{2}\right)|}$