Question

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

Answer 1

Let the slope of the required line be m.

Then, its equation is $\frac{y-2}{x-3}=m$

Given line is $x-2y=3\Rightarrow y=\frac{1}{2}x-\frac{3}{2}$

Clearly, the slope of this line is $\frac{1}{2}$

It is given that the angle between (i) and (ii) is 45${}^{\circ }$.

$\left|\begin{array}{c}\frac{m-\frac{1}{2}}{1+\frac{1}{2}m}\end{array}\right|$ = $\text{tan}{45}^{\circ }$

$[\text{using tan}\theta =\left|\begin{array}{c}\frac{m_2-m_1}{1+m_1{m}_2}\end{array}\right|]$

$\iff \frac{2m-1}{2+m}=1\text{or}\frac{2m-1}{2+m}=-1\iff m=3\text{or}m=\frac{-1}{3}$

$\therefore$ the required equation is $\frac{y-2}{x-3}=3\text{or}\frac{y-2}{x-3}=\frac{-1}{3}$

i.e.,$3x-y-7=0orx+3y-9=0$