# Distance between two parallel lines

The distance between two parallel lines in the plane is the minimum distance between any two points.

## Formula and proof

Because the lines are parallel, the perpendicular distance between them is a constant, so it does not matter which point is chosen to measure the distance. Given the equations of two non-vertical parallel lines

$y=mx+b_{1}\,$
$y=mx+b_{2}\,,$

the distance between the two lines is the distance between the two intersection points of these lines with the perpendicular line

$y=-x/m\,.$

This distance can be found by first solving the linear systems

${\begin{cases}y=mx+b_{1}\\y=-x/m\,,\end{cases}}$

and

${\begin{cases}y=mx+b_{2}\\y=-x/m\,,\end{cases}}$

to get the coordinates of the intersection points. The solutions to the linear systems are the points

$\left(x_{1},y_{1}\right)\ =\left({\frac {-b_{1}m}{m^{2}+1}},{\frac {b_{1}}{m^{2}+1}}\right)\,,$

and

$\left(x_{2},y_{2}\right)\ =\left({\frac {-b_{2}m}{m^{2}+1}},{\frac {b_{2}}{m^{2}+1}}\right)\,.$

The distance between the points is

$d={\sqrt {\left({\frac {b_{1}m-b_{2}m}{m^{2}+1}}\right)^{2}+\left({\frac {b_{2}-b_{1}}{m^{2}+1}}\right)^{2}}}\,,$

which reduces to

$d={\frac {|b_{2}-b_{1}|}{\sqrt {m^{2}+1}}}\,.$

When the lines are given by

$ax+by+c_{1}=0\,$
$ax+by+c_{2}=0,\,$

the distance between them can be expressed as

$d={\frac {|c_{2}-c_{1}|}{\sqrt {a^{2}+b^{2}}}}.$