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.