# Find the perpendicular distance from the origin to the line joining the points

Question:

Find the perpendicular distance from the origin to the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$

Solution:

The equation of the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ is given by

$y-\sin \theta=\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}(x-\cos \theta)$

$y(\cos \phi-\cos \theta)-\sin \theta(\cos \phi-\cos \theta)=x(\sin \phi-\sin \theta)-\cos \theta(\sin \phi-\sin \theta)$

$x(\sin \theta-\sin \phi)+y(\cos \phi-\cos \theta)+\cos \theta \sin \phi-\cos \theta \sin \theta-\sin \theta \cos \phi+\sin \theta \cos \theta=0$

$x(\sin \theta-\sin \phi)+y(\cos \phi-\cos \theta)+\sin (\phi-\theta)=0$

$A x+B y+C=0$, where $A=\sin \theta-\sin \phi, B=\cos \phi-\cos \theta$, and $C=\sin (\phi-\theta)$

It is known that the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by $d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$.

Therefore, the perpendicular distance $(d)$ of the given line from point $\left(x_{1}, y_{1}\right)=(0,0)$ is

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

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

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

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

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

$=\frac{|\sin (\phi-\theta)|}{\sqrt{2\left(2 \sin ^{2}\left(\frac{\phi-\theta}{2}\right)\right)}}$

$=\frac{|\sin (\phi-\theta)|}{\left|2 \sin \left(\frac{\phi-\theta}{2}\right)\right|}$