Find the equations of the line through the point (3, 2)
which make an angle of ${45}^{\circ }$ with the line $x-2y-3=0$

Solution :-

REF.Image

Given $CD\Rightarrow x-2y=3$

Let slope of AB be m

slope of given liner $=\frac{-A}{B}=-\frac{1}{2}=\frac{1}{2}$

Now $tan\theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=tan{45}^{\circ }$

$\left|\frac{\frac{1}{2}-m}{1+1/2}\right|=1$

$\Rightarrow |\frac{1}{2}-m|=1+\frac{1}{2}$

$+(\frac{1}{2}-m)=1+\frac{1}{2}$ or $-(\frac{1}{2}-m)=1+\frac{1}{2}$

Solving we get m $-1/3$ and $m=3$

Now when $m=-1/3$

equation of line $y-y_1=m(x-x_1)$

$y-2=\frac{-1}{3}(x-3)$

Required line $\Rightarrow 3y-6=-x+3\Rightarrow x+3y=9$

or when $m=3$

$\therefore y-y_1=n(x-x_1)$

$y-2=3(x-3)$

Required line $\Rightarrow 3x-y-7=0$