Find the perpendicular distance from the origin of the line joini…

Question

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

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Answer

Equation of the line joining points $(\cos \theta ,\sin \theta )$ and $(\cos \phi ,\sin \phi )$.
$y - \sin \theta = \frac{{\sin \phi - \sin \theta }}{{\cos \phi - \cos \theta }}(x - \cos \theta )$
$ \Rightarrow (\cos \phi - \cos \theta )y - \sin \theta \cos \phi $$ + \sin \theta \cos \theta $
$ = (\sin \phi - \sin \theta )x - \sin \phi \cos \theta $$ + \sin \theta \cos \theta $
$ \Rightarrow (\sin \phi - \sin \phi )x - (\cos \phi - \cos \theta )$$y - \sin \phi \cos \theta + \sin \theta \cos \phi = 0$
$ \Rightarrow (\sin \phi - \sin \phi )x - (\cos \phi - \cos \theta )y$$ + \sin (\theta - \phi ) = 0$
Now perpendicular distance from (0, 0) to the given line is
$ = \left| {\frac{\begin{gathered} (\sin \phi - \sin \theta ) \times 0 - (\cos \phi - \cos \theta \times 0 \hfill \\ + \sin (\theta - \phi ) \hfill \\ \end{gathered} }{{\sqrt {{{(\sin \phi - \sin \theta )}^2} + {{(\cos \phi - \cos \theta )}^2}} }}} \right|$
$ = \left| {\frac{{\sin (\theta - \phi )}}{{\sqrt {{{\sin }^2}\phi + {{\sin }^2}\theta - 2\sin \phi \sin \theta + {{\cos }^2}\phi + {{\cos }^2}\theta - 2\cos \phi \cos \theta } }}} \right|$
$ = \left| {\frac{{\sin (\theta - \phi )}}{{\sqrt {2 - 2(\cos \theta \cos \phi + \sin \theta \sin \theta )} }}} \right|$
$ = \left| {\frac{{\sin (\theta - \phi )}}{{\sqrt {2[1 - \cos (\theta - \phi )]} }}} \right| = \left| {\frac{{\sin (\theta - \phi )}}{{\sqrt {2\left[ {2{{\sin }^2}\left( {\frac{{\theta - \phi }}{2}} \right)} \right]} }}} \right|$
$=\frac{{|\sin (\theta - \phi )|}}{{\left| {2\sin \left( {\frac{{\theta - \phi }}{2}} \right)} \right|}}$.