Point-Line Distance--2-D

Given a line $ax+by+c = 0$ and a point $(x_0,y_0$), in slope-intercept form, the equation of the line is

$$y = - {a\over b} x - {c\over b},$$ (1)

so the line has Slope $-a/b$. Points on the line have the vector coordinates

$$\left[{\matrix{x\cr -{a\over b}x-{c\over d}\cr}}\right] = \l... ...r d}\cr}}\right]-{1\over b}\left[{\matrix{-b\cr a\cr}}\right].$$ (2)

Therefore, the Vector

$$\left[{\matrix{-b\cr a\cr}}\right]$$ (3)

is Parallel to the line, and the Vector

$${\bf v} =\left[{\matrix{a\cr b\cr}}\right]$$ (4)

is Perpendicular to it. Now, a Vector from the point to the line is given by

$${\bf r} = \left[{\matrix{x-x_0\cr y-y_0\cr}}\right].$$ (5)

Projecting ${\bf r}$ onto ${\bf v}$,

$$\vert{\rm proj}_{\bf v}{\bf r}\vert = {\vert{\bf v}\cdot {\bf r}\vert\over \vert{\bf v}\vert} = \vert{\hat {\bf v}\cdot{\bf r}}\vert = {\vert a(x-x_0)+b(y-y_0)\vert\over \sqrt{a^2+b^2}}$$

$$= {\vert ax+by-ax_0-by_0\vert\over \sqrt{a^2+b^2}}$$

$$= {\vert ax_0+by_0+c\vert\over \sqrt{a^2+b^2}}.$$ (6)

If the line is represented by the endpoints of a Vector $(x_1, y_1)$ and $(x_2, y_2)$, then the Perpendicular Vector is

$$\bf v= \left[{\matrix{y_2-y_1\cr -(x_2-x_1)\cr}}\right]$$ (7)

$$\hat{\bf v}= {1\over s}\left[{\matrix{y_2-y_1\cr -(x_2-x_1)\cr}}\right],$$ (8)

where

$$s=\vert{\bf v}\vert=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},$$ (9)

so the distance is

$$d=\vert{\hat {\bf v}\cdot{\bf r}}\vert ={\vert(y_2-y_1)(x_0-x_1)-(x_2-x_1)(y_0-y_1)\vert\over s}.$$ (10)

The distance from a point ($x_1$,$y_1$) to the line $y=a+bx$ can be computed using Vector algebra. Let ${\bf L}$ be a Vector in the same direction as the line

$${\bf L} = \left[\begin{array}{c}x\\ a+bx\end{array}\right]-\left[\begin{array}{c}0\\ a\end{array}\right]=\left[\begin{array}{c}x\\ bx\end{array}\right]$$ (11)

$$\hat {\bf L} = {1\over \sqrt{b^2+1}} \left[\begin{array}{c}1\\ b\end{array}\right].$$ (12)

A given point on the line is

$${\bf x}=\left[{\matrix{x_1\cr y_1\cr}}\right]-\left[{\matrix{0\cr -a\cr}}\right] = \left[{\matrix{x_1\cr y_1-a\cr}}\right],$$ (13)

so the point-line distance is

$${\bf r} = ({\bf x}\cdot \hat{\bf L})\hat{\bf L}-{\bf x}$$

$$= {1\over 1+b^2}\left({\left[\begin{array}{c}x_1\\ y_1-a\end{array... ...nd{array}\right] -\left[\begin{array}{c}x_1\\ y_1-a\end{array}\right]\nonumber$$

$$= {x_1+b(y_1-a)\over 1+b^2}\left[\begin{array}{c}1\\ b\end{array}\right]-\left[\begin{array}{c}x_1\\ y_1-a\end{array}\right]$$

$$= {1\over 1+b^2}\left[\begin{array}{c}b(y_1-a)-b^2x_1\\ bx_1+b^2y_1-ab^2-y_1+a-b^2y_1+ab^2\end{array}\right]$$

$$= {1\over 1+b^2} \left[\begin{array}{c}b[(y_1-a)-bx_1]\\ -[(y_1-a)-bx_1]\end{array}\right]$$

$$= {y_1-(a+bx_1)\over 1+b^2} \left[\begin{array}{c}b\\ -1\end{array}\right].$$ (14)

Therefore,

$$d = \vert{\bf r}\vert = {\vert y_1-(a+bx_1)\vert\over 1+b^2} \sqrt{1+b^2} = {\vert y_1-(a+bx_1)\vert\over\sqrt{1+b^2}}.$$ (15)

This result can also be obtained much more simply by noting that the Perpendicular distance is just $\cos\theta$ times the vertical distance $\vert y_1-(a+bx_1)\vert$. But the Slope $b$ is just $\tan\theta$, so

$$\sin^2\theta +\cos ^2\theta =1 \Rightarrow \tan^2\theta +1={1\over\cos^2\theta},$$ (16)

and

$$\cos\theta={1\over \sqrt{1+\tan^2\theta}}={1\over\sqrt{1+b^2}}.$$ (17)

The Perpendicular distance is then

$$d={\vert y_1-(a+bx_1)\vert\over\sqrt{1+b^2}},$$ (18)

the same result as before.

See also Line, Point, Point-Line Distance--3-D