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

Let the slope of the required line be $m_{1}$.
The given line is $x-2y=3$,which can be written as $y=\frac{1}{2}x-\frac{3}{2}$.
Thus,the slope of the given line is $m_{2}=\frac{1}{2}$.
The angle $\theta$ between the two lines is $45^{\circ}$. The formula for the angle between two lines with slopes $m_{1}$ and $m_{2}$ is $\tan \theta = \left| \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right|$.
Substituting the values: $\tan 45^{\circ} = \left| \frac{\frac{1}{2}-m_{1}}{1+m_{1}(\frac{1}{2})} \right|$.
$1 = \left| \frac{1-2m_{1}}{2+m_{1}} \right|$.
This gives two cases:
Case $I$: $\frac{1-2m_{1}}{2+m_{1}} = 1$ $\Rightarrow 1-2m_{1} = 2+m_{1}$ $\Rightarrow 3m_{1} = -1$ $\Rightarrow m_{1} = -\frac{1}{3}$.
Case $II$: $\frac{1-2m_{1}}{2+m_{1}} = -1$ $\Rightarrow 1-2m_{1} = -2-m_{1}$ $\Rightarrow m_{1} = 3$.
For $m_{1} = 3$,the equation of the line passing through $(3,2)$ is $y-2 = 3(x-3)$ $\Rightarrow y-2 = 3x-9$ $\Rightarrow 3x-y = 7$.
For $m_{1} = -\frac{1}{3}$,the equation of the line passing through $(3,2)$ is $y-2 = -\frac{1}{3}(x-3)$ $\Rightarrow 3y-6 = -x+3$ $\Rightarrow x+3y = 9$.
Thus,the equations of the lines are $3x-y=7$ and $x+3y=9$.