distance from point to a line
The distance from a point P with coordinates (xp,yp)∈ℝ2 to the line with equation ax+by+c=0 is given by |axp+byp+c|/√a2+b2.
Proof Every point x,y on the line is at some distance √(x-xp)2+(y-yp)2 from P. What we need to find is the minimum such distance. Our problem is
min(x-xp)2+(y-yp)2 |
subject to
ax+by+c=0 |
This problem is solvable using the Lagrange multiplier method. We minimize
(x-xp)2+(y-yp)2+λ(ax+by+c) |
Calculating the derivatives with respect to x,y and λ and setting them to zero we get three equations:
2x-2xp+λa=0 | (1) | ||
2y-2yp+λb=0 | (2) | ||
2ax+2by+2c=0 | (3) |
Solving these leads to xp-x=aaxp+byp+ca2+b2 and yp-y=baxp+byp+ca2+b2. We can now substitute these expressions into √(x-xp)2+(y-yp)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 |