Distance between a line and a point

In many cases, we will need to find the shortest distance between a line a point.

This distance is on the perpendicular of the line. This means that we need to get the intersection point and calculate the distance between the two points.

For instance, if the point is $P$ and the lines intersect at $I$ , we need the distance $PI$ .

Distance between point $P(x_0,y_0)$ and line $ax+by+c=0$

The point is $P(x_0,y_0)$ is on the line perpendicular to the given line $ax+by+c=0$

The product of the slopes is $m_1m_2=-1$

We start by putting $ax+by+c=0$ in the slope-intercept form

$ax+by+c=0$
$by=-ax-c$
$y=-\frac{a}{b}x-\frac{c}{b}$

That means:
$m_1=-\frac{a}{b}$

Now we have to find $m_2$ from $m_1$ :

$m_1m_2=-1$

$-\frac{a}{b}m_2=-1$

$\frac{a}{b}m_2=1$

$m_2=\frac{b}{a}$ after we multiply both sides by $\frac{b}{a}$

The perpendicular is a line passing through $P(x_0,y_0)$ with a slope of $m_2=\frac{b}{a}$

We get for the perpendicular:

$y-y_0=\frac{b}{a}(x-x_0)$

$y=\frac{b}{a}x-\frac{b}{a}x_0+y_0$

When the two lines meet, they have corresponding coordinates $x$ and $y$ .

We get:

$y=-\frac{a}{b}x-\frac{c}{b}$

$y=\frac{b}{a}x-\frac{b}{a}x_0+y_0$

$-\frac{a}{b}x-\frac{c}{b}=\frac{b}{a}x-\frac{b}{a}x_0+y_0$

$x(\frac{b}{a}+\frac{a}{b})=-\frac{c}{b}+\frac{b}{a}x_0-y_0$

$x(\frac{a^{2}+b^{2}}{ab})=\frac{-ac+b^{2}x_0-aby_0}{ab}$

Now we multiply both sides by $ab$

$x(a^{2}+b^{2})=-ac+b^{2}x_0-aby_0$

Dividing by $a^{2}+b^{2}$

$\boxed{x=\frac{-ac+b^{2}x_0-aby_0}{a^{2}+b^{2}}}$

To find $y$ we use any of the two lines and plug in $x$ expression.

From:

$y=-\frac{a}{b}x-\frac{c}{b}$

We get

(1) $\begin{equation*}\begin{split}$y&=-\frac{a}{b}(\frac{-ac+b^{2}x_0-aby_0}{a^{2}+b^{2}})-\frac{c}{b}\\& =\frac{a^{2}c-ab^{2}x_0+a^{2}by_0}{b(a^{2}+b^{2})}-\frac{c(a^{2}+b^{2})}{b(a^{2}+b^{2})}\\& =\frac{a^{2}c-a^{2}c-b^{2}c-ab^{2}x_0+a^{2}by_0}{b(a^{2}+b^{2})}\\& =\frac{-b^{2}c-ab^{2}x_0+a^{2}by_0}{b(a^{2}+b^{2})}\\&=\frac{a^{2}y_0-bc-abx_0}{a^{2}+b^{2}}\end{split}\end{equation*}$

Finally:

$\boxed{y=\frac{a^{2}y_0-bc-abx_0}{a^{2}+b^{2}}}$

Now we have the two points, all we need is to calculate the distance:

$P=(x_0, y_0)$

$I=(\frac{ac+b^{2}x_0-aby_0}{a^{2}+b^{2}},\frac{a^{2}y_0-bc-abx_0}{a^{2}+b^{2}})$

Distance $IP$ :

(2) $\begin{equation*}\begin{split}d&=\sqrt{(x_0-\frac{-ac+b^{2}x_0-aby_0}{a^{2}+b^{2}})^{2}+ (y_0-\frac{a^{2}y_0-bc-abx_0}{a^{2}+b^{2}})^{2}}\\& =\sqrt{(\frac{a^{2}x_0+ac+b^{2}x_0-b^{2}x_0+aby_0}{a^{2}+b^{2}})^{2}+(\frac{a^{2}y_0-a^{2}y_0+b^{2}y_0+bc+abx_0}{a^{2}+b^{2}})^{2}}\\& =\sqrt{(\frac{a^{2}x_0+ac+aby_0}{a^{2}+b^{2}})^{2}+(\frac{b^{2}y_0+bc+abx_0}{a^{2}+b^{2}})^{2}}\\& =\sqrt{a^{2}(\frac{ax_0+by_0+c}{a^{2}+b^{2}})^{2}+b^{2}(\frac{ax_0+by_0+c}{a^{2}+b^{2}})^{2}}\\& =\sqrt{(a^{2}+b^{2})(\frac{ax_0+by_0+c}{a^{2}+b^{2}})^{2}}\\& =\sqrt{\frac{(ax_0+by_0+c)^{2}}{a^{2}+b^{2}}}\end{split}\end{equation*}$