If the angle between two lines is $\frac{\pi }{4}$and slope of one of the lines is $\frac{1}{2}$, find the slope of the other line.

Solution: Let $m_1$ and $m_2$ are the slopes of these two lines and $\theta$ is the angle between these two lines. Then,
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
Here, $\theta =\frac{\pi }{4}\text{and}{m}_1=\frac{1}{2}$
$\therefore \tan \left(\frac{\pi }{4}\right)=\left|\frac{\frac{1}{2}-m_2}{1+\frac{1}{2}{m}_2}\right|$
$\Rightarrow 1=\left|\frac{\frac{1}{2}-m_2}{1+\frac{1}{2}{m}_2}\right|$
$\Rightarrow \frac{\frac{1}{2}-m_2}{1+\frac{1}{2}{m}_2}=1\text{or}\frac{\frac{1}{2}-m_2}{1+\frac{1}{2}{m}_2}=-1$
$\Rightarrow \left(\frac{1}{2}-m_2\right)=\left(1+\frac{1}{2}{m}_2\right)\text{or}\left(\frac{1}{2}-m_2\right)=-\left(1+\frac{1}{2}{m}_2\right)$
$\Rightarrow \frac{3}{2}{m}_2=-\frac{1}{2}\text{or}-\frac{1}{2}{m}_2=-\frac{3}{2}$
$\Rightarrow m_2=-\frac{1}{3}\text{or}{m}_2=3$
So, slope of other line will be $-\frac{1}{3}$ or $3$.