Find the perpendicular distance between the pair of parallel lines 3x-4y+12 = 0 and 3x-4y+2=0
How do i start ??
With your question there are two different ways to find the answer:
1. Choose any arbitrary point on the 1rst line; determine the equation of a line perpendicular to the 1rst line passing through this point; calculate the intersection point of the perpendicular line and the 2nd line; calculate the distance between the choosen point and the intersection point.
2. The equations of the two lines are given in normal form:
$\displaystyle 3x-4y+12=0~\implies~(3, -4)\cdot (x, y) +12 = 0$ with
$\displaystyle (3, -4)$ indicating the perpendicular direction to the line and
12 indicating the perpendicular distance of the origin to the line.
If the normal vector is a unit vector then the distance of origin to the line is measured in units too. The difference between these two distances must be the distance between the two parallels:
$\displaystyle |(3, -4)| = 5$ Therefore the equation of the two lines become:
$\displaystyle \frac35x-\frac45y+\frac{12}5=0\ and\ \frac35x-\frac45y+\frac{2}5=0$
Therefore the distance between the parallels is:
$\displaystyle d = \frac{12}5 - \frac25 = 2$