## Distance from a Point to a Line

2016 Dec 1

Derive a formula for the distance from a point to a line in the cartesian coordinate plane.

The distance from any point to a line is defined as the shortest distance between that point and any point on the line. The line segment connecting a point to its closest point on a line is always perpendicular to the line. Let the line be the line $$y=mx+b$$ with angle of elevation with the x-axis $$\theta$$ and the point be the point $$A(x_0,y_0)$$. Let $$\overline{AB}$$ be the perpendicular line segment from $$A$$ to $$y=mx+b$$, so that the distance $$d$$ is equal to the length of $$\overline{AB}$$. Let us also drop a vertical line segment $$\overline{AC}$$ from $$A$$ to $$C$$, the point on $$y=mx+b$$ directly above or below it. Finally, construct a line parallel to the x-axis through the point $$C$$. Now we have something that looks like this:

Now we can see that we have constructed a right triangle. Note that the angle that $$y=mx+b$$ makes with the horizontal line through $$C$$ is also $$\theta$$, since the horizontal line through $$C$$ is parallel to the x-axis and the line $$y-mx+b$$ acts as a transversal. Then, since $$\overline{AC}$$ is perpendicular to the horizontal line through $$C$$, $$\angle{ACB}=90^{\circ}-\theta$$. Since $$\theta=tan^{-1}(m)$$,

$\angle{ACB}=90^{\circ}-tan^{-1}(m)=tan^{-1}(\frac{1}{m})$

Also, since the ordinate of $$A$$ is $$y_0$$ and the ordinate of $$C$$ is $$mx_0+b$$,

$\overline{AC}=\mid y_0-mx_0-b\mid$

This is all we need to calculate $$d$$ using right triangle trigonometry:

$sin(\angle{ACB})=\frac{d}{\mid y_0-mx_0-b\mid}$
$sin(tan^{-1}(\frac{1}{m}))=\frac{d}{\mid y_0-mx_0-b\mid}$
$\frac{1}{\sqrt{m^2+1}}=\frac{d}{\mid y_0-mx_0-b\mid}$
$d=\frac{\mid y_0-mx_0-b\mid}{\sqrt{m^{2}+1}}$

Which gives us the distance between the point and the line.