Question

Find the distance between the parallel lines $3x-4y+9=0$ and $6x-8y-15=0$.

Answer 1

Putting $y=0$ in $3x-4y+9=0$, we get $x=-3$.

Thus, $(-3,0)$ is a point on the line $3x-4y+9=0$.

Length of the perpendicular from $(-3,0)$ to $6x-8y-15=0$ is given by

$d=\frac{|-3\times 6-8\times 0-15|}{\sqrt{6^2+(-8{)}^2}}=\frac{33}{10}$