Distance between two parallel lines: Difference between revisions

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The '''[[Distancedistance]]''' between two [[Parallel (geometry)|parallel]] [[Line (geometry)|lines]]''' in the plane is the perpendicular distance from a [[Point (geometry)|point]] of theon firstone line to the anotherother line.

== Formula and proof ==

Because athe lines are parallel linethe isperpendicular adistance linebetween thatthem hasis ana equalconstant, distanceso withit thedoes oppositenot line,matter therewhich point is achosen uniqueto distance betweenmeasure the two parallel linesdistance. Given the equations of two non-vertical parallel lines

:<math>y = mx+b_1\,</math>

:<math>y = mx+b_2\,,</math>

the distance between the two lines can be found by solving the linear systems

the distance between the two lines is the distance between the two intercepts of these lines with the perpendicular line

:<math>y = -x/m \, ,</math>

theThis distance between the two lines can be found by first solving 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 intercept 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>

:<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>

The distance between the points is

:<math>d = \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

:<math>ax+by+c_1=0\,</math>

:<math>ax+by+c_2=0,\,</math>

theirthe distance between them can be expressed as

:<math>d = \frac{|c_2-c_1|}{\sqrt {a^2+b^2}}.</math>