distance from point to a line

The distance from a point P with coordinates $(x_{p},y_{p})\in\mathbb{R}^{2}$ to the line with equation $ax+by+c=0$ is given by $|ax_{p}+by_{p}+c|/\sqrt{a^{2}+b^{2}}$ .

Proof Every point $x, y$ on the line is at some distance $\sqrt{(x-x_{p})^{2}+(y-y_{p})^{2}}$ from P. What we need to find is the minimum such distance. Our problem is

\min(x-x_{p})^{2}+(y-y_{p})^{2}

subject to

ax+by+c=0

This problem is solvable using the Lagrange multiplier method. We minimize

(x-x_{p})^{2}+(y-y_{p})^{2}+\lambda(ax+by+c)

Calculating the derivatives with respect to $x, y$ and $\lambda$ and setting them to zero we get three equations:

	$\displaystyle 2x-2x_{p}+\lambda a=0$		(1)
	$\displaystyle 2y-2y_{p}+\lambda b=0$		(2)
	$\displaystyle 2ax+2by+2c=0$		(3)

Solving these leads to $x_{p}-x=a\frac{ax_{p}+by_{p}+c}{a^{2}+b^{2}}$ and $y_{p}-y=b\frac{ax_{p}+by_{p}+c}{a^{2}+b^{2}}$ . We can now substitute these expressions into $\sqrt{(x-x_{p})^{2}+(y-y_{p})^{2}}$ and we get (after some simplification) the desired result.

Title	distance from point to a line
Canonical name	DistanceFromPointToALine
Date of creation	2013-03-22 15:24:30
Last modified on	2013-03-22 15:24:30
Owner	acastaldo (8031)
Last modified by	acastaldo (8031)
Numerical id	7
Author	acastaldo (8031)
Entry type	Result
Classification	msc 51N20
Related topic	DistanceOfNonParallelLines
Related topic	DistanceBetweenTwoLinesInR3
Related topic	Envelope
Related topic	AngleBisectorAsLocus