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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    $\small \frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }$.

Views

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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