Find the equation of the lines through the point $(3,2)$ which make an angle of $45{}^{\circ }$ with the line $x-2y=3$.

Let the slope of the required line be $m_1$.
The given line can be written as $y=\frac{1}{2}x-\frac{3}{2}$, which is of the form $y=mx+c$
$\therefore$ slope of the given line $=m_2=\frac{1}{2}$
It is given that the angle between the required line and line $x-2y=3$ is ${45}^{\circ }$
We know that if $\theta$ is the acute angle between lines $l_1$ and $l_2$ with slopes $m_1$ and $m_2$ respectively, then $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$\therefore \tan {45}^{\circ }=\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$\Rightarrow 1=\left|\frac{\frac{1}{2}-m_1}{1+\frac{m_1}{2}}\right|$
$\Rightarrow 1=\left|\frac{\left(\frac{1-2m_1}{2}\right)}{\frac{2+m_1}{2}}\right|$
$\Rightarrow 1=\pm \left|\frac{1-2m_1}{2+m_1}\right|$
$\Rightarrow m_1=-\frac{1}{3}or{m}_1=3$
case 1; $m_1=3$
The equation of the line passing through $(3,2)$ and having a slope of $3$ is
$y-2=3(x-3)\Rightarrow 3x-y=7$
case 2: $m_1=-\frac{1}{3}$
The equation of the line passing through $(3,2)$ and having a slope of $-\frac{1}{3}$ is.
$y-2=-\frac{1}{3}(x-3)$
$\Rightarrow 3y-6=x+3\Rightarrow x+3y=9$