Question

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

Answer 1

Let the equation of the line passing through the origin be $y=m_{1}x$

If this line makes an angle of $\theta$ with line $y=mx+c$, then angle $\theta$ is given by

$\tan \theta =\left|\frac{m_1-m}{1+m_{1}m}\right|$

$\Rightarrow \tan \theta =\left|\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right|$

$\Rightarrow \tan \theta =\pm \frac{\frac{y}{x}-m}{1+\frac{y}{x}m}$

$\Rightarrow \tan \theta =\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}$

or $\tan \theta =-\left|\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right|$

$\tan \theta =\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}$

Case1:

$\tan \theta =\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}$

$\Rightarrow \tan \theta +\frac{y}{x}m\tan \theta =\frac{y}{x}-m$

$\Rightarrow m+\tan \theta =\frac{y}{x}(1-m\tan \theta )$

$\Rightarrow \frac{y}{x}=\frac{m+\tan \theta }{1-m\tan \theta }$

Case 2:

$\therefore \tan \theta =-\left\{\frac{\frac{y}{x}-m}{1+\frac{y}{x}m}\right\}$

$\Rightarrow \tan \theta +\frac{y}{x}m\tan \theta =-\frac{y}{x}+m$

$\Rightarrow \frac{y}{x}(1+m\tan \theta )=m-\tan \theta$

$\Rightarrow \frac{y}{x}=\frac{m-\tan \theta }{1+m\tan \theta }$

Therefore, the required line is given by $\frac{y}{x}=\frac{m\mp \tan \theta }{1\pm m\tan \theta }$