Question

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

Answer 1

Equation of line passing through $(cos\theta ,sin\theta )$ and $(cos\varphi ,sin\varphi )$ is

$y-sin\varphi =\left(\frac{sin\varphi ,-sin\theta }{cos\varphi -cos\theta }\right)(x-cos\varphi )$

$y-sin\varphi =\left(\frac{2cos\frac{\theta +\varphi }{2}sin\frac{\varphi -\theta }{2}}{-2sin\frac{\theta +\varphi }{2}sin\frac{\varphi -\theta }{2}}\right)(x-cos\varphi )$

$y-sin\varphi =-cot\left(\frac{\theta +\varphi }{2}\right)(x-cos\varphi )$

$xcot\left(\frac{\theta +\varphi }{2}\right)+y-sin\varphi -cos\varphi cot\left(\frac{\theta +\varphi }{2}\right)=0$

Distance of this line from origin,

$=\left|\frac{a{x}_{1}by+c}{\sqrt{a^2+b^2}}\right|$

$\left|\frac{0+0-sin\varphi -cos\varphi cot\left(\frac{\theta +\varphi }{2}\right)}{\sqrt{(cos{\left(\frac{\theta +\varphi }{2}\right)}^2+1)}}\right|$

$\left|\frac{sin\varphi +cos\varphi cot\left(\frac{\theta +\varphi }{2}\right)}{cosec\left(\frac{\theta +\varphi }{2}\right)}\right|$

$=sin\varphi sin\left(\frac{\theta +\varphi }{2}\right)+cos\varphi cot\left(\frac{\theta +\varphi }{2}\right)$

$=cos(\frac{\theta +\varphi }{2}-\varphi )$

$=cos\left(\frac{\theta +\varphi -2\varphi }{2}\right)$

$D=cos\left(\frac{\theta -\varphi }{2}\right)$