Question

Find the joint equation of lines passing through the point $(3,2)$ and one is parallel to $x-2y=2$ and other is perpendicular to $y=3$.

Answer 1

The line parallel to the line $x-2y=2$ is $x-2y=k$ which passes through $(3,2)$ then we get, $3-2.2=k$ or, $k=-1$.

So the equation of the line is $x-2y=-\mathrm{1.....}\left(1\right)$

Again the line perpendicular to $y=3$ is $x=k$, which passes through $(3,2)$ then we've, $x=\mathrm{3.....}\left(2\right)$

Now joint equation represented by $\left(1\right)$ and $\left(2\right)$ is,

$(x-3)(x-2y+1)=0$.