Question

Find the co-ordinates of the points on each line but at a unit distance from the other line. Their equations are $3x-4y+1=0$ and $8x+6y+1=0$.

Answer 1

The two lines are perpendicular and meet at $O$ $(-\frac{2}{10},\frac{1}{10})$ and $\tan \theta =\frac{3}{4}$ for $AB$.
$\therefore \cos \theta =4/5,\sin \theta =3/5$
$x+\frac{2}{10}y-\frac{1}{10}$
$\therefore AB$ is $\frac{10}{4/5}=\frac{10}{3/5}=\begin{array}{c}1\\ forA\end{array},\begin{array}{c}-1\\ forB\end{array}$
$\therefore A$ is $(\frac{4}{5}-\frac{2}{10},\frac{3}{5}+\frac{1}{10})$
and $B$ is $(-\frac{4}{5}-\frac{2}{10},-\frac{3}{5}+\frac{1}{10})$
or $A$ is $(\frac{3}{5},\frac{7}{10})$ and $B$ is $(-1,-\frac{1}{2})$
It can be verified that distance of these two points from line $CD$ $8x+6y+1=0$ is each unity.
Similarly $C$ and $D$ can be found on $8x+6y+1=0$, $\tan \theta =-4/3$
so that $\cos \theta =-3/5,$ $\sin \theta =4/5$.
$CD$ is $\frac{x+\frac{2}{10}}{-3/5}=\frac{y-\frac{1}{10}}{4/5}=\begin{array}{c}1\\ C\end{array},\begin{array}{c}-1\\ D\end{array}$
$\therefore C$ is $(-\frac{8}{10},\frac{9}{10})$ and $D$ is $(\frac{4}{10},-\frac{7}{10})$.