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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find perpendicular distance from the origin to the line joining the points $ (\cos \theta, \sin \theta) $ and $(\cos \phi, \sin \phi)$

$\begin {array} {1 1} (A)\;\large\frac{|\sin (\phi- \theta)|}{| \sin \Large\frac{(\phi - \theta)}{2}|} & \quad (B)\;\large\frac{\sin (\phi- \theta)}{ \sin \Large\frac{(\phi - \theta)}{2}} \\ (C)\;\large\frac{|\sin (\phi- \theta)|}{|2 \sin \Large\frac{(\phi - \theta)}{2}|} & \quad (D)\;\large\frac{\sin (\phi- \theta)}{2 \sin \Large\frac{(\phi - \theta)}{2}} \end {array}$

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  • Equation of a line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $ \large\frac{y-y_1}{y_2-y_1}$$ = \large\frac{x-x_1}{x_2-x_1}$
  • Equation of a line having slope $m$ and passing through the point $(x_1, y_1)$ is $ y-y_1=m(x-x_1)$
  • Perpendicular distance $d$ of a line $Ax+By+C=0$ from a point $(x_1, y_1)$ is given by $ d = \large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
The given points are $ (\cos \theta, \sin \theta )$ and $( \cos \phi, \sin \phi)$
Hence the equation of the line joining these points is
$ \large\frac{x-\cos \theta}{\cos \phi - \cos \theta}$$ = \large\frac{y- \sin \theta}{\sin \phi - \sin \theta}$
$ \Rightarrow (x-\cos \theta ) ( \sin \phi - \sin \theta ) = (y- \sin \theta ) ( \cos \phi - \cos \theta)$
$ \Rightarrow x(\sin \phi - \sin \theta ) -y( \cos \phi - \cos \theta )- \cos \theta \sin \phi+ \cos \theta \not{\sin} \not{\theta} + \sin \theta \cos \phi - \not{\sin} \not{\theta} \cos \theta=0$
But $ \cos \phi \sin \theta - \cos \theta \sin \phi = \sin ( \phi - \theta)$
$ \therefore x( \sin \phi - \sin \theta )-y( \cos \phi - \cos \theta ) + \sin ( \phi - \theta )=0$
Comparing this with the equation $Ax+By+C = 0$
$ A = \sin \theta - \sin \phi , \: B = \cos \phi - \cos \theta $ and $ C = \sin ( \phi - \theta)$
Perpendicular distance $d$ of the line $Ax+By+C=0$ from a point $(x_1, y_1)$ is $ d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
Substituting the values we get,
hence $d =\large\frac{ |(\sin \theta - \sin \phi ) (0)+ ( \cos \phi - \cos \theta )(0)+\sin ( \phi - \theta )|}{\sqrt{(\sin \theta - \sin \phi)^2+(\cos \phi - \cos \theta )^2}}$
$ = \large\frac{|\sin \theta - \phi|}{\sqrt{\sin^2 \theta + \sin^2 \phi - 2 \sin \theta \sin \phi + \cos^2\theta+ \cos^2 \phi - 2\cos \theta \cos \phi}}$
$ = \large\frac{ | \sin ( \theta - \phi )|}{ \sqrt{ ( \sin^2 \theta + \cos^2 \theta )+( \sin^2 \phi + \cos^2 \phi )-2( \sin \theta \sin \phi + \cos \theta \cos \phi)}}$
$ = \large\frac{| \sin ( \phi - \theta)|}{\sqrt{1+1-2\cos ( \phi - \theta)}}$
$ = \large\frac{| \sin ( \phi - \theta )|}{\sqrt{ 2 ( 1-\cos ( \phi - \theta )}}$
Hint $ \because 1-\cos \theta = 2 \sin^2 \large\frac{ \theta}{2}$
$ = \large\frac{ |\sin ( \phi - \theta )|}{ \sqrt{2(2 \sin^2 \Large\frac{( \phi - \theta)}{2}}}$
$ = \large\frac{ | \sin ( \phi - \theta )|}{ \bigg| 2 \sin \bigg( \Large\frac{\phi - \theta }{2} \bigg) \bigg|}$
answered May 20, 2014 by thanvigandhi_1
 

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