Step 1 :
Angle made with the line $x-2y=3$ is $45^{\circ}$
Hence $ \tan 45^{\circ}=1$
(i.e) $m_1=1$
Slope of the line $ x-2y=3$ is $ m_2 = -\bigg( \large\frac{1}{-2} \bigg)$
$m_2 = \large\frac{1}{2}$
$ \therefore \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
$ = \bigg| \large\frac{1-\Large\frac{1}{2} }{1+1 \times \Large\frac{1}{2} } \bigg|$ or $ \bigg| \large\frac{-1-\Large\frac{1}{2}}{1+(-1) \bigg( \Large\frac{1}{2} \bigg)} \bigg|$
$ = \bigg| \large\frac{\Large\frac{1}{2}}{\Large\frac{3}{2}} \bigg|$$ = \large\frac{1}{3}$
$ = \bigg|\large\frac{- \Large\frac{3}{2}}{\Large\frac{1}{2}} \bigg|$$ = 3$
Hence the slope of the required line are 3 and $ -\large\frac{1}{3}$
It passes through the point (3,2)
Hence the equations of the lines are where $m = 3$
$y-2 = 3(x-3)$
$ \Rightarrow (y-2)=3(x-3)$
$ \Rightarrow y-2=3x-9$
or $3x-y=7$--------(1)
where $m = -\large\frac{1}{3}$
$y-2=-\large\frac{1}{3}$$ (x-3)$
$3y-6=-x+3$
$ \Rightarrow x+3y=9$---------(2)
Hence the required equations of the line are
$3x-y=7 $ or $ x+3y=9$