## 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.