Let $P Q$ be the line $y=m x+c$, whose slope is $\mathrm{m}$ The line $P O$ makes an angle $\theta$ with $P Q$. Let slope of $P O$ be $m_1$.


$\therefore \quad \tan \theta=\pm \frac{m_1-m}{1+m_1 m}$
Take $+$ ve sign, $\tan \theta=\frac{m_1-m}{1+m_1 m}$
or $\tan \theta+m_1 m \tan \theta=m_1-m$
$\Rightarrow \quad m+\tan \theta=m_1-m_1 m \tan \theta$
$\Rightarrow \quad m+\tan \theta=m_1(1-m \tan \theta)$
$\Rightarrow \quad m_1=\frac{m+\tan \theta}{1-m \tan \theta}$
Take-ve sign, $\tan \theta=-\frac{\mathrm{m}_1-\mathrm{m}}{1+\mathrm{m}_1 \mathrm{~m}}$
$\Rightarrow \quad \tan \theta+m_1 m \tan \theta=-m_1+m$
$\Rightarrow \quad m_1(1+m \tan \theta)=m-\tan \theta$
$\therefore \quad m_1=\frac{m-\tan \theta}{1+m \tan \theta}$
$\therefore \quad$ Equation of the line $O P$ is $y=\mathrm{m}_1 x$
or $\quad \frac{y}{x}=m_1 \quad \therefore \quad \frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$