Find the equation of the lines through the point (3, 2)

Solution:

Let the slope of the required line be m1.

The given line can be written as $y=\frac{1}{2} x-\frac{3}{2}$, which is of the form $y=m x+c$

$\therefore$ Slope of the given line $=m_{2}=\frac{1}{2}$

It is given that the angle between the required line and line $x-2 y=3$ is $45^{\circ}$.

We know that if $\theta$ isthe acute angle between lines $I_{1}$ and $I_{2}$ with slopes $m_{1}$ and $m_{2}$ respectively, then $\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$.

$\therefore \tan 45^{\circ}=\frac{\left|m_{1}-m_{2}\right|}{1+m_{1} m_{2}}$

$\Rightarrow 1=\left|\begin{array}{l}\frac{1}{2}-m_{1} \\ 1+\frac{m_{1}}{2}\end{array}\right|$

$\Rightarrow 1=\left|\frac{\left(\frac{1-2 m_{1}}{2}\right)}{\frac{2+m_{1}}{2}}\right|$

$\Rightarrow 1=\left|\frac{1-2 m_{1}}{2+m_{1}}\right|$

$\Rightarrow 1=\pm\left(\frac{1-2 m_{1}}{2+m_{1}}\right)$

$\Rightarrow 1=\frac{1-2 m_{1}}{2+m_{1}}$ or $1=-\left(\frac{1-2 m_{1}}{2+m_{1}}\right)$

$\Rightarrow 2+m_{1}=1-2 m_{1}$ or $2+m_{1}=-1+2 m_{1}$

$\Rightarrow m_{1}=-\frac{1}{3}$ or $m_{1}=3$

Case I: m1 = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y – 2 = 3 (x – 3)

y – 2 = 3x – 9

3x – y = 7

Case II: $m_{1}=-\frac{1}{3}$

The equation of the line passing through $(3,2)$ and having a slope of $-\frac{1}{3}$ is:

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

$3 y-6=-x+3$

$x+3 y=9$

Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.