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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Show that the equation of the line passing through the origin and making an angle $ \theta $ with the line $y=mx+c$ is $ \large\frac{y}{x}$$ = \large\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$

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  • Equation of the line having intercepts $a$ and $b$ on the axes is $ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$
Let the equation of the line passing through the origin be $y=m_1x \Rightarrow m_1= \large\frac{y}{x}$
Let then line make an angle $ \theta$ with the line $y=mx+c$
$ \therefore \tan \theta = \bigg| \large\frac{m_1-m}{1+m_1m} \bigg|$
$ \Rightarrow \tan \theta = \bigg| \large\frac{\Large\frac{y}{x}-m}{1+ \Large\frac{y}{x}.m} \bigg|$
$ \Rightarrow \tan \theta = \large\frac{\bigg( \Large\frac{y}{x}-m \bigg)}{1+ \Large\frac{y}{x}m }$ or $ \tan \theta = \large\frac{-\bigg( \large\frac{y}{x}-m \bigg)}{1+\large\frac{y}{x}m}$
Case (i)
If $ \tan \theta = \large\frac{\bigg( \large\frac{y}{x}-m \bigg)}{1+\large\frac{y}{x}m}$
$ \Rightarrow \tan \theta \bigg( 1+ \large\frac{y}{x}$$m \bigg) $$ = \large\frac{y}{x}$$-m$
$ \Rightarrow \tan \theta + \tan \theta. \large\frac{y}{x}$$m= \large\frac{y}{x}$$-m$
$ \Rightarrow m + \tan \theta = \large\frac{y}{x}$$ (1- m \: \tan \theta )$
$ \Rightarrow \large\frac{y}{x}$$ = \large\frac{m+ \tan \theta}{1-m \tan \theta}$
Case (ii)
When $ \tan \theta = \large\frac{-\bigg( \large\frac{y}{x}-m\bigg)}{1+ \large\frac{y}{x}m}$
$ \Rightarrow \tan \theta \bigg( 1+ \large\frac{y}{x}$$m \bigg) = - \bigg( \large\frac{y}{x}$$-m \bigg)$
$ \Rightarrow \tan \theta + \tan \theta \large\frac{y}{x}$$m = -\large\frac{y}{x}$$+m$
$ \therefore \large\frac{y}{x}$$ ( 1+ m \tan \theta)=m-\tan \theta$
$ \Rightarrow \large\frac{y}{x}$$= \large\frac{m-\tan \theta}{1+m \tan \theta}$
Hence the required line is given by
$ \large\frac{y}{x} = \large\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$
Hence proved.
answered May 21, 2014 by thanvigandhi_1
 

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