Slopes of the lines through the point (3, 2) and making an angle ${45}^0$ with the line $x-2y=3$ are

Let 'm' be the slope of the line passing through the point (3, 2) 

Let $\theta ={45}^0,$

and slope of $x-2y-3=0 is \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Let $m_1=m, m_2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$$\begin{gathered} Tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ \Rightarrow Tan{45}^0=\left|\frac{m-{\displaystyle \raisebox{1ex}{$ 1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+m\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}\right| \\ \Rightarrow 1 = \left|\frac{2m-1}{2+m}\right| \\ \Rightarrow \frac{2m-1}{2+m}=\pm 1 \end{gathered}$$

$$\begin{gathered} \therefore \frac{2m-1}{2+m}=1 or \frac{2m-1}{2+m}=-1 \\ \Rightarrow 2m-1 =2+m or 2m-1 =-2-m \\ \Rightarrow m=3 or 3m=-1 \\ m=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$