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Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x=m+-tan θ/1-+m tanθ.

Q : 13     Show that the equation of the line passing through the origin and making an angle
                \small \theta with the line  \small y=mx+c is    \small \frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }.
 

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Slope of line \small y=mx+c is m
Let  the slope of other line is m'
It is given that both the line makes  an angle  \small \theta with each other
Therefore,
\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
\tan \theta = \left | \frac{m-m'}{1+mm'} \right |
\mp(1+mm')\tan \theta =(m-m')
\mp\tan \theta +m'(\mp m\tan\theta+1)= m
m'= \frac{m\pm \tan \theta}{1\mp m\tan \theta}
Now, equation of line passing through origin (0,0) and with slope  \frac{m\pm \tan \theta}{1\mp m\tan \theta}  is
(y-0)=\frac{m\pm \tan \theta}{1\mp m\tan \theta}(x-0)
\frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta}
Hence proved 

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