Editing Parallel (geometry) (section)

==== Distance between two parallel lines ====
{{Main|Distance between two parallel lines}}

Because parallel lines in a Euclidean plane are [[equidistant]] there is a unique distance between the two parallel lines. Given the equations of two non-vertical, non-horizontal parallel lines,
:<math>y = mx+b_1\,</math>
:<math>y = mx+b_2\,,</math>
the distance between the two lines can be found by locating two points (one on each line) that lie on a common perpendicular to the parallel lines and calculating the distance between them. Since the lines have slope ''m'', a common perpendicular would have slope −1/''m'' and we can take the line with equation ''y'' = −''x''/''m'' as a common perpendicular. Solve the linear systems
:<math>\begin{cases}
y = mx+b_1 \\
y = -x/m
\end{cases}</math>
and
:<math>\begin{cases}
y = mx+b_2 \\
y = -x/m
\end{cases}</math>
to get the coordinates of the points. The solutions to the linear systems are the points
:<math>\left( x_1,y_1 \right)\ = \left( \frac{-b_1m}{m^2+1},\frac{b_1}{m^2+1} \right)\,</math>
and
:<math>\left( x_2,y_2 \right)\ = \left( \frac{-b_2m}{m^2+1},\frac{b_2}{m^2+1} \right).</math>
These formulas still give the correct point coordinates even if the parallel lines are horizontal (i.e., ''m'' = 0). The distance between the points is
:<math>d = \sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2} = \sqrt{\left(\frac{b_1m-b_2m}{m^2+1}\right)^2 + \left(\frac{b_2-b_1}{m^2+1}\right)^2}\,,</math>
which reduces to
:<math>d = \frac{|b_2-b_1|}{\sqrt{m^2+1}}\,.</math>

When the lines are given by the general form of the equation of a line (horizontal and vertical lines are included):
:<math>ax+by+c_1=0\,</math>
:<math>ax+by+c_2=0,\,</math>
their distance can be expressed as
:<math>d = \frac{|c_2-c_1|}{\sqrt {a^2+b^2}}.</math>