Shortest distance between a point and a line segment

Here is the simplest complete code in Javascript.

x, y is your target point and x1, y1 to x2, y2 is your line segment.

UPDATED: fix for 0 length line problem from comments.

function pDistance(x, y, x1, y1, x2, y2) { var A = x - x1; var B = y - y1; var C = x2 - x1; var D = y2 - y1; var dot = A * C + B * D; var len_sq = C * C + D * D; var param = -1; if (len_sq != 0) //in case of 0 length line param = dot / len_sq; var xx, yy; if (param < 0) { xx = x1; yy = y1; } else if (param > 1) { xx = x2; yy = y2; } else { xx = x1 + param * C; yy = y1 + param * D; } var dx = x - xx; var dy = y - yy; return Math.sqrt(dx * dx + dy * dy); } 

UPDATED: Kotlin version

fun getDistance(x: Double, y: Double, x1: Double, y1: Double, x2: Double, y2: Double): Double { val a = x - x1 val b = y - y1 val c = x2 - x1 val d = y2 - y1 val lenSq = c * c + d * d val param = if (lenSq != .0) { //in case of 0 length line val dot = a * c + b * d dot / lenSq } else { -1.0 } val (xx, yy) = when { param < 0 -> x1 to y1 param > 1 -> x2 to y2 else -> x1 + param * c to y1 + param * d } val dx = x - xx val dy = y - yy return hypot(dx, dy) } 

This is an implementation made for FINITE LINE SEGMENTS, not infinite lines like most other functions here seem to be (that's why I made this).

Implementation of theory by Paul Bourke.

Python:

def dist(x1, y1, x2, y2, x3, y3): # x3,y3 is the point px = x2-x1 py = y2-y1 norm = px*px + py*py u = ((x3 - x1) * px + (y3 - y1) * py) / float(norm) if u > 1: u = 1 elif u < 0: u = 0 x = x1 + u * px y = y1 + u * py dx = x - x3 dy = y - y3 # Note: If the actual distance does not matter, # if you only want to compare what this function # returns to other results of this function, you # can just return the squared distance instead # (i.e. remove the sqrt) to gain a little performance dist = (dx*dx + dy*dy)**.5 return dist 

AS3:

public static function segmentDistToPoint(segA:Point, segB:Point, p:Point):Number { var p2:Point = new Point(segB.x - segA.x, segB.y - segA.y); var something:Number = p2.x*p2.x + p2.y*p2.y; var u:Number = ((p.x - segA.x) * p2.x + (p.y - segA.y) * p2.y) / something; if (u > 1) u = 1; else if (u < 0) u = 0; var x:Number = segA.x + u * p2.x; var y:Number = segA.y + u * p2.y; var dx:Number = x - p.x; var dy:Number = y - p.y; var dist:Number = Math.sqrt(dx*dx + dy*dy); return dist; } 

Java

private double shortestDistance(float x1,float y1,float x2,float y2,float x3,float y3) { float px=x2-x1; float py=y2-y1; float temp=(px*px)+(py*py); float u=((x3 - x1) * px + (y3 - y1) * py) / (temp); if(u>1){ u=1; } else if(u<0){ u=0; } float x = x1 + u * px; float y = y1 + u * py; float dx = x - x3; float dy = y - y3; double dist = Math.sqrt(dx*dx + dy*dy); return dist; } 

In my own question thread how to calculate shortest 2D distance between a point and a line segment in all cases in C, C# / .NET 2.0 or Java? I was asked to put a C# answer here when I find one: so here it is, modified from http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static :

//Compute the dot product AB . BC private double DotProduct(double[] pointA, double[] pointB, double[] pointC) { double[] AB = new double[2]; double[] BC = new double[2]; AB[0] = pointB[0] - pointA[0]; AB[1] = pointB[1] - pointA[1]; BC[0] = pointC[0] - pointB[0]; BC[1] = pointC[1] - pointB[1]; double dot = AB[0] * BC[0] + AB[1] * BC[1]; return dot; } //Compute the cross product AB x AC private double CrossProduct(double[] pointA, double[] pointB, double[] pointC) { double[] AB = new double[2]; double[] AC = new double[2]; AB[0] = pointB[0] - pointA[0]; AB[1] = pointB[1] - pointA[1]; AC[0] = pointC[0] - pointA[0]; AC[1] = pointC[1] - pointA[1]; double cross = AB[0] * AC[1] - AB[1] * AC[0]; return cross; } //Compute the distance from A to B double Distance(double[] pointA, double[] pointB) { double d1 = pointA[0] - pointB[0]; double d2 = pointA[1] - pointB[1]; return Math.Sqrt(d1 * d1 + d2 * d2); } //Compute the distance from AB to C //if isSegment is true, AB is a segment, not a line. double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC, bool isSegment) { double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB); if (isSegment) { double dot1 = DotProduct(pointA, pointB, pointC); if (dot1 > 0) return Distance(pointB, pointC); double dot2 = DotProduct(pointB, pointA, pointC); if (dot2 > 0) return Distance(pointA, pointC); } return Math.Abs(dist); } 

I'm @SO not to answer but ask questions so I hope I don't get million down votes for some reasons but constructing critic. I just wanted (and was encouraged) to share somebody else's ideas since the solutions in this thread are either with some exotic language (Fortran, Mathematica) or tagged as faulty by somebody. The only useful one (by Grumdrig) for me is written with C++ and nobody tagged it faulty. But it's missing the methods (dot etc.) that are called.

For anyone interested, here's a trivial conversion of Joshua's Javascript code to Objective-C:

- (double)distanceToPoint:(CGPoint)p fromLineSegmentBetween:(CGPoint)l1 and:(CGPoint)l2 { double A = p.x - l1.x; double B = p.y - l1.y; double C = l2.x - l1.x; double D = l2.y - l1.y; double dot = A * C + B * D; double len_sq = C * C + D * D; double param = dot / len_sq; double xx, yy; if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) { xx = l1.x; yy = l1.y; } else if (param > 1) { xx = l2.x; yy = l2.y; } else { xx = l1.x + param * C; yy = l1.y + param * D; } double dx = p.x - xx; double dy = p.y - yy; return sqrtf(dx * dx + dy * dy); } 

I needed this solution to work with MKMapPoint so I will share it in case someone else needs it. Just some minor change and this will return the distance in meters :

- (double)distanceToPoint:(MKMapPoint)p fromLineSegmentBetween:(MKMapPoint)l1 and:(MKMapPoint)l2 { double A = p.x - l1.x; double B = p.y - l1.y; double C = l2.x - l1.x; double D = l2.y - l1.y; double dot = A * C + B * D; double len_sq = C * C + D * D; double param = dot / len_sq; double xx, yy; if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) { xx = l1.x; yy = l1.y; } else if (param > 1) { xx = l2.x; yy = l2.y; } else { xx = l1.x + param * C; yy = l1.y + param * D; } return MKMetersBetweenMapPoints(p, MKMapPointMake(xx, yy)); } 

In F#, the distance from the point c to the line segment between a and b is given by:

let pointToLineSegmentDistance (a: Vector, b: Vector) (c: Vector) = let d = b - a let s = d.Length let lambda = (c - a) * d / s let p = (lambda |> max 0.0 |> min s) * d / s (a + p - c).Length 

The vector d points from a to b along the line segment. The dot product of d/s with c-a gives the parameter of the point of closest approach between the infinite line and the point c. The min and max function are used to clamp this parameter to the range 0..s so that the point lies between a and b. Finally, the length of a+p-c is the distance from c to the closest point on the line segment.

Example use:

pointToLineSegmentDistance (Vector(0.0, 0.0), Vector(1.0, 0.0)) (Vector(-1.0, 1.0)) 

In Mathematica

It uses a parametric description of the segment, and projects the point into the line defined by the segment. As the parameter goes from 0 to 1 in the segment, if the projection is outside this bounds, we compute the distance to the corresponding enpoint, instead of the straight line normal to the segment.

Clear["Global`*"]; distance[{start_, end_}, pt_] := Module[{param}, param = ((pt - start).(end - start))/Norm[end - start]^2; (*parameter. the "." here means vector product*) Which[ param < 0, EuclideanDistance[start, pt], (*If outside bounds*) param > 1, EuclideanDistance[end, pt], True, EuclideanDistance[pt, start + param (end - start)] (*Normal distance*) ] ]; 

Plotting result:

Plot3D[distance[{{0, 0}, {1, 0}}, {xp, yp}], {xp, -1, 2}, {yp, -1, 2}] 

Plot those points nearer than a cutoff distance:

Contour Plot:

Hey, I just wrote this yesterday. It's in Actionscript 3.0, which is basically Javascript, though you might not have the same Point class.

//st = start of line segment //b = the line segment (as in: st + b = end of line segment) //pt = point to test //Returns distance from point to line segment. //Note: nearest point on the segment to the test point is right there if we ever need it public static function linePointDist( st:Point, b:Point, pt:Point ):Number { var nearestPt:Point; //closest point on seqment to pt var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product var bLenSq:Number = dot( b, b ); //Segment length squared if( keyDot <= 0 ) //pt is "behind" st, use st { nearestPt = st } else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz) { nearestPt = st.add(b); } else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point { var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB ); nearestPt = st.add(partOfB); } var dist:Number = (pt.subtract(nearestPt)).length; return dist; } 

Also, there's a pretty complete and readable discussion of the problem here: notejot.com

One line solution using arctangents:

The idea is to move A to (0, 0) and rotate triangle clockwise to make C lay on X axis, when this happen, By will be the distance.

1. a angle = Atan(Cy - Ay, Cx - Ax);
2. b angle = Atan(By - Ay, Bx - Ax);
3. AB length = Sqrt( (Bx - Ax)^2 + (By - Ay)^2 )
4. By = Sin ( bAngle - aAngle) * ABLength

C#

public double Distance(Point a, Point b, Point c) { // normalize points Point cn = new Point(c.X - a.X, c.Y - a.Y); Point bn = new Point(b.X - a.X, b.Y - a.Y); double angle = Math.Atan2(bn.Y, bn.X) - Math.Atan2(cn.Y, cn.X); double abLength = Math.Sqrt(bn.X*bn.X + bn.Y*bn.Y); return Math.Sin(angle)*abLength; } 

One line C# (to be converted to SQL)

double distance = Math.Sin(Math.Atan2(b.Y - a.Y, b.X - a.X) - Math.Atan2(c.Y - a.Y, c.X - a.X)) * Math.Sqrt((b.X - a.X) * (b.X - a.X) + (b.Y - a.Y) * (b.Y - a.Y)) 

For the lazy, here's my Objective-C port of @Grumdrig's solution above:

CGFloat sqr(CGFloat x) { return x*x; } CGFloat dist2(CGPoint v, CGPoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); } CGFloat distanceToSegmentSquared(CGPoint p, CGPoint v, CGPoint w) { CGFloat l2 = dist2(v, w); if (l2 == 0.0f) return dist2(p, v); CGFloat t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2; if (t < 0.0f) return dist2(p, v); if (t > 1.0f) return dist2(p, w); return dist2(p, CGPointMake(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y))); } CGFloat distanceToSegment(CGPoint point, CGPoint segmentPointV, CGPoint segmentPointW) { return sqrtf(distanceToSegmentSquared(point, segmentPointV, segmentPointW)); } 

Couldn't resist coding it in python :)

from math import sqrt, fabs def pdis(a, b, c): t = b[0]-a[0], b[1]-a[1] # Vector ab dd = sqrt(t[0]**2+t[1]**2) # Length of ab t = t[0]/dd, t[1]/dd # unit vector of ab n = -t[1], t[0] # normal unit vector to ab ac = c[0]-a[0], c[1]-a[1] # vector ac return fabs(ac[0]*n[0]+ac[1]*n[1]) # Projection of ac to n (the minimum distance) print pdis((1,1), (2,2), (2,0)) # Example (answer is 1.414) 

Ditto for fortran :)

real function pdis(a, b, c) real, dimension(0:1), intent(in) :: a, b, c real, dimension(0:1) :: t, n, ac real :: dd t = b - a ! Vector ab dd = sqrt(t(0)**2+t(1)**2) ! Length of ab t = t/dd ! unit vector of ab n = (/-t(1), t(0)/) ! normal unit vector to ab ac = c - a ! vector ac pdis = abs(ac(0)*n(0)+ac(1)*n(1)) ! Projection of ac to n (the minimum distance) end function pdis program test print *, pdis((/1.0,1.0/), (/2.0,2.0/), (/2.0,0.0/)) ! Example (answer is 1.414) end program test 

Here is a more complete spelling out of Grumdrig's solution. This version also returns the closest point itself.

#include "stdio.h" #include "math.h" class Vec2 { public: float _x; float _y; Vec2() { _x = 0; _y = 0; } Vec2( const float x, const float y ) { _x = x; _y = y; } Vec2 operator+( const Vec2 &v ) const { return Vec2( this->_x + v._x, this->_y + v._y ); } Vec2 operator-( const Vec2 &v ) const { return Vec2( this->_x - v._x, this->_y - v._y ); } Vec2 operator*( const float f ) const { return Vec2( this->_x * f, this->_y * f ); } float DistanceToSquared( const Vec2 p ) const { const float dX = p._x - this->_x; const float dY = p._y - this->_y; return dX * dX + dY * dY; } float DistanceTo( const Vec2 p ) const { return sqrt( this->DistanceToSquared( p ) ); } float DotProduct( const Vec2 p ) const { return this->_x * p._x + this->_y * p._y; } }; // return minimum distance between line segment vw and point p, and the closest point on the line segment, q float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q ) { const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt if ( distSq == 0.0 ) { // v == w case (*q) = v; return v.DistanceTo( p ); } // consider the line extending the segment, parameterized as v + t (w - v) // we find projection of point p onto the line // it falls where t = [(p-v) . (w-v)] / |w-v|^2 const float t = ( p - v ).DotProduct( w - v ) / distSq; if ( t < 0.0 ) { // beyond the v end of the segment (*q) = v; return v.DistanceTo( p ); } else if ( t > 1.0 ) { // beyond the w end of the segment (*q) = w; return w.DistanceTo( p ); } // projection falls on the segment const Vec2 projection = v + ( ( w - v ) * t ); (*q) = projection; return p.DistanceTo( projection ); } float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY ) { Vec2 q; float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q ); (*qX) = q._x; (*qY) = q._y; return distance; } void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY ) { float qX; float qY; float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY ); printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n", segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY ); } void TestDistanceFromLineSegmentToPoint() { TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 ); TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 ); TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 ); TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 ); TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 ); TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 ); } 

I'm assuming you want to find the shortest distance between the point and a line segment; to do this, you need to find the line (lineA) which is perpendicular to your line segment (lineB) which goes through your point, determine the intersection between that line (lineA) and your line which goes through your line segment (lineB); if that point is between the two points of your line segment, then the distance is the distance between your point and the point you just found which is the intersection of lineA and lineB; if the point is not between the two points of your line segment, you need to get the distance between your point and the closer of two ends of the line segment; this can be done easily by taking the square distance (to avoid a square root) between the point and the two points of the line segment; whichever is closer, take the square root of that one.

Consider this modification to Grumdrig's answer above. Many times you'll find that floating point imprecision can cause problems. I'm using doubles in the version below, but you can easily change to floats. The important part is that it uses an epsilon to handle the "slop". In addition, you'll many times want to know WHERE the intersection happened, or if it happened at all. If the returned t is < 0.0 or > 1.0, no collision occurred. However, even if no collision occurred, many times you'll want to know where the closest point on the segment to P is, and thus I use qx and qy to return this location.

double PointSegmentDistanceSquared( double px, double py, double p1x, double p1y, double p2x, double p2y, double& t, double& qx, double& qy) { static const double kMinSegmentLenSquared = 0.00000001; // adjust to suit. If you use float, you'll probably want something like 0.000001f static const double kEpsilon = 1.0E-14; // adjust to suit. If you use floats, you'll probably want something like 1E-7f double dx = p2x - p1x; double dy = p2y - p1y; double dp1x = px - p1x; double dp1y = py - p1y; const double segLenSquared = (dx * dx) + (dy * dy); if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared) { // segment is a point. qx = p1x; qy = p1y; t = 0.0; return ((dp1x * dp1x) + (dp1y * dp1y)); } else { // Project a line from p to the segment [p1,p2]. By considering the line // extending the segment, parameterized as p1 + (t * (p2 - p1)), // we find projection of point p onto the line. // It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2 t = ((dp1x * dx) + (dp1y * dy)) / segLenSquared; if (t < kEpsilon) { // intersects at or to the "left" of first segment vertex (p1x, p1y). If t is approximately 0.0, then // intersection is at p1. If t is less than that, then there is no intersection (i.e. p is not within // the 'bounds' of the segment) if (t > -kEpsilon) { // intersects at 1st segment vertex t = 0.0; } // set our 'intersection' point to p1. qx = p1x; qy = p1y; // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if // we were doing PointLineDistanceSquared, then qx would be (p1x + (t * dx)) and qy would be (p1y + (t * dy)). } else if (t > (1.0 - kEpsilon)) { // intersects at or to the "right" of second segment vertex (p2x, p2y). If t is approximately 1.0, then // intersection is at p2. If t is greater than that, then there is no intersection (i.e. p is not within // the 'bounds' of the segment) if (t < (1.0 + kEpsilon)) { // intersects at 2nd segment vertex t = 1.0; } // set our 'intersection' point to p2. qx = p2x; qy = p2y; // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if // we were doing PointLineDistanceSquared, then qx would be (p1x + (t * dx)) and qy would be (p1y + (t * dy)). } else { // The projection of the point to the point on the segment that is perpendicular succeeded and the point // is 'within' the bounds of the segment. Set the intersection point as that projected point. qx = p1x + (t * dx); qy = p1y + (t * dy); } // return the squared distance from p to the intersection point. Note that we return the squared distance // as an optimization because many times you just need to compare relative distances and the squared values // works fine for that. If you want the ACTUAL distance, just take the square root of this value. double dpqx = px - qx; double dpqy = py - qy; return ((dpqx * dpqx) + (dpqy * dpqy)); } } 

Grumdrig's C++/JavaScript implementation was very useful to me, so I have provided a Python direct port that I am using. The complete code is here.

class Point(object): def __init__(self, x, y): self.x = float(x) self.y = float(y) def square(x): return x * x def distance_squared(v, w): return square(v.x - w.x) + square(v.y - w.y) def distance_point_segment_squared(p, v, w): # Segment length squared, |w-v|^2 d2 = distance_squared(v, w) if d2 == 0: # v == w, return distance to v return distance_squared(p, v) # Consider the line extending the segment, parameterized as v + t (w - v). # We find projection of point p onto the line. # It falls where t = [(p-v) . (w-v)] / |w-v|^2 t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / d2; if t < 0: # Beyond v end of the segment return distance_squared(p, v) elif t > 1.0: # Beyond w end of the segment return distance_squared(p, w) else: # Projection falls on the segment. proj = Point(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y)) # print proj.x, proj.y return distance_squared(p, proj) 

Here it is using Swift

 /* Distance from a point (p1) to line l1 l2 */ func distanceFromPoint(p: CGPoint, toLineSegment l1: CGPoint, and l2: CGPoint) -> CGFloat { let A = p.x - l1.x let B = p.y - l1.y let C = l2.x - l1.x let D = l2.y - l1.y let dot = A * C + B * D let len_sq = C * C + D * D var param : CGFloat = -1 if len_sq != 0 { param = dot / len_sq } var xx, yy: CGFloat if param < 0 || (l1.x == l2.x && l1.y == l2.y) { xx = l1.x yy = l1.y } else if param > 1 { xx = l2.x yy = l2.y } else { xx = l1.x + param * C yy = l1.y + param * D } let dx = p.x - xx let dy = p.y - yy return sqrt(dx * dx + dy * dy) } 

And now my solution as well...... (Javascript)

It is very fast because I try to avoid any Math.pow functions.

As you can see, at the end of the function I have the distance of the line.

code is from the lib http://www.draw2d.org/graphiti/jsdoc/#!/example

/** * Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2) * A simple hit test. * * @return {boolean} * @static * @private * @param {Number} coronaWidth the accepted corona for the hit test * @param {Number} X1 x coordinate of the start point of the line * @param {Number} Y1 y coordinate of the start point of the line * @param {Number} X2 x coordinate of the end point of the line * @param {Number} Y2 y coordinate of the end point of the line * @param {Number} px x coordinate of the point to test * @param {Number} py y coordinate of the point to test **/ graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1, X2, Y2, px, py) { // Adjust vectors relative to X1,Y1 // X2,Y2 becomes relative vector from X1,Y1 to end of segment X2 -= X1; Y2 -= Y1; // px,py becomes relative vector from X1,Y1 to test point px -= X1; py -= Y1; var dotprod = px * X2 + py * Y2; var projlenSq; if (dotprod <= 0.0) { // px,py is on the side of X1,Y1 away from X2,Y2 // distance to segment is length of px,py vector // "length of its (clipped) projection" is now 0.0 projlenSq = 0.0; } else { // switch to backwards vectors relative to X2,Y2 // X2,Y2 are already the negative of X1,Y1=>X2,Y2 // to get px,py to be the negative of px,py=>X2,Y2 // the dot product of two negated vectors is the same // as the dot product of the two normal vectors px = X2 - px; py = Y2 - py; dotprod = px * X2 + py * Y2; if (dotprod <= 0.0) { // px,py is on the side of X2,Y2 away from X1,Y1 // distance to segment is length of (backwards) px,py vector // "length of its (clipped) projection" is now 0.0 projlenSq = 0.0; } else { // px,py is between X1,Y1 and X2,Y2 // dotprod is the length of the px,py vector // projected on the X2,Y2=>X1,Y1 vector times the // length of the X2,Y2=>X1,Y1 vector projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2); } } // Distance to line is now the length of the relative point // vector minus the length of its projection onto the line // (which is zero if the projection falls outside the range // of the line segment). var lenSq = px * px + py * py - projlenSq; if (lenSq < 0) { lenSq = 0; } return Math.sqrt(lenSq)<coronaWidth; }; 

C#

Adapted from @Grumdrig

public static double MinimumDistanceToLineSegment(this Point p, Line line) { var v = line.StartPoint; var w = line.EndPoint; double lengthSquared = DistanceSquared(v, w); if (lengthSquared == 0.0) return Distance(p, v); double t = Math.Max(0, Math.Min(1, DotProduct(p - v, w - v) / lengthSquared)); var projection = v + t * (w - v); return Distance(p, projection); } public static double Distance(Point a, Point b) { return Math.Sqrt(DistanceSquared(a, b)); } public static double DistanceSquared(Point a, Point b) { var d = a - b; return DotProduct(d, d); } public static double DotProduct(Point a, Point b) { return (a.X * b.X) + (a.Y * b.Y); } 

Matlab code, with built-in "self test" if they call the function with no arguments:

function r = distPointToLineSegment( xy0, xy1, xyP ) % r = distPointToLineSegment( xy0, xy1, xyP ) if( nargin < 3 ) selfTest(); r=0; else vx = xy0(1)-xyP(1); vy = xy0(2)-xyP(2); ux = xy1(1)-xy0(1); uy = xy1(2)-xy0(2); lenSqr= (ux*ux+uy*uy); detP= -vx*ux + -vy*uy; if( detP < 0 ) r = norm(xy0-xyP,2); elseif( detP > lenSqr ) r = norm(xy1-xyP,2); else r = abs(ux*vy-uy*vx)/sqrt(lenSqr); end end function selfTest() %#ok<*NASGU> disp(['invalid args, distPointToLineSegment running (recursive) self-test...']); ptA = [1;1]; ptB = [-1;-1]; ptC = [1/2;1/2]; % on the line ptD = [-2;-1.5]; % too far from line segment ptE = [1/2;0]; % should be same as perpendicular distance to line ptF = [1.5;1.5]; % along the A-B but outside of the segment distCtoAB = distPointToLineSegment(ptA,ptB,ptC) distDtoAB = distPointToLineSegment(ptA,ptB,ptD) distEtoAB = distPointToLineSegment(ptA,ptB,ptE) distFtoAB = distPointToLineSegment(ptA,ptB,ptF) figure(1); clf; circle = @(x, y, r, c) rectangle('Position', [x-r, y-r, 2*r, 2*r], ... 'Curvature', [1 1], 'EdgeColor', c); plot([ptA(1) ptB(1)],[ptA(2) ptB(2)],'r-x'); hold on; plot(ptC(1),ptC(2),'b+'); circle(ptC(1),ptC(2), 0.5e-1, 'b'); plot(ptD(1),ptD(2),'g+'); circle(ptD(1),ptD(2), distDtoAB, 'g'); plot(ptE(1),ptE(2),'k+'); circle(ptE(1),ptE(2), distEtoAB, 'k'); plot(ptF(1),ptF(2),'m+'); circle(ptF(1),ptF(2), distFtoAB, 'm'); hold off; axis([-3 3 -3 3]); axis equal; end end 

coded in t-sql

the point is (@px, @py) and the line segment runs from (@ax, @ay) to (@bx, @by)

create function fn_sqr (@NumberToSquare decimal(18,10)) returns decimal(18,10) as begin declare @Result decimal(18,10) set @Result = @NumberToSquare * @NumberToSquare return @Result end go create function fn_Distance(@ax decimal (18,10) , @ay decimal (18,10), @bx decimal(18,10), @by decimal(18,10)) returns decimal(18,10) as begin declare @Result decimal(18,10) set @Result = (select dbo.fn_sqr(@ax - @bx) + dbo.fn_sqr(@ay - @by) ) return @Result end go create function fn_DistanceToSegmentSquared(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) returns decimal(18,10) as begin declare @l2 decimal(18,10) set @l2 = (select dbo.fn_Distance(@ax, @ay, @bx, @by)) if @l2 = 0 return dbo.fn_Distance(@px, @py, @ax, @ay) declare @t decimal(18,10) set @t = ((@px - @ax) * (@bx - @ax) + (@py - @ay) * (@by - @ay)) / @l2 if (@t < 0) return dbo.fn_Distance(@px, @py, @ax, @ay); if (@t > 1) return dbo.fn_Distance(@px, @py, @bx, @by); return dbo.fn_Distance(@px, @py, @ax + @t * (@bx - @ax), @ay + @t * (@by - @ay)) end go create function fn_DistanceToSegment(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) returns decimal(18,10) as begin return sqrt(dbo.fn_DistanceToSegmentSquared(@px, @py , @ax , @ay , @bx , @by )) end go --example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1) select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1) --result = 2.2360679775 --example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1) select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1) --result = 2.4961508830 --example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1) select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1) --result = 0.0000000000 

Looks like just about everyone else on StackOverflow has contributed an answer (23 answers so far), so here's my contribution for C#. This is mostly based on the answer by M. Katz, which in turn is based on the answer by Grumdrig.

 public struct MyVector { private readonly double _x, _y; // Constructor public MyVector(double x, double y) { _x = x; _y = y; } // Distance from this point to another point, squared private double DistanceSquared(MyVector otherPoint) { double dx = otherPoint._x - this._x; double dy = otherPoint._y - this._y; return dx * dx + dy * dy; } // Find the distance from this point to a line segment (which is not the same as from this // point to anywhere on an infinite line). Also returns the closest point. public double DistanceToLineSegment(MyVector lineSegmentPoint1, MyVector lineSegmentPoint2, out MyVector closestPoint) { return Math.Sqrt(DistanceToLineSegmentSquared(lineSegmentPoint1, lineSegmentPoint2, out closestPoint)); } // Same as above, but avoid using Sqrt(), saves a new nanoseconds in cases where you only want // to compare several distances to find the smallest or largest, but don't need the distance public double DistanceToLineSegmentSquared(MyVector lineSegmentPoint1, MyVector lineSegmentPoint2, out MyVector closestPoint) { // Compute length of line segment (squared) and handle special case of coincident points double segmentLengthSquared = lineSegmentPoint1.DistanceSquared(lineSegmentPoint2); if (segmentLengthSquared < 1E-7f) // Arbitrary "close enough for government work" value { closestPoint = lineSegmentPoint1; return this.DistanceSquared(closestPoint); } // Use the magic formula to compute the "projection" of this point on the infinite line MyVector lineSegment = lineSegmentPoint2 - lineSegmentPoint1; double t = (this - lineSegmentPoint1).DotProduct(lineSegment) / segmentLengthSquared; // Handle the two cases where the projection is not on the line segment, and the case where // the projection is on the segment if (t <= 0) closestPoint = lineSegmentPoint1; else if (t >= 1) closestPoint = lineSegmentPoint2; else closestPoint = lineSegmentPoint1 + (lineSegment * t); return this.DistanceSquared(closestPoint); } public double DotProduct(MyVector otherVector) { return this._x * otherVector._x + this._y * otherVector._y; } public static MyVector operator +(MyVector leftVector, MyVector rightVector) { return new MyVector(leftVector._x + rightVector._x, leftVector._y + rightVector._y); } public static MyVector operator -(MyVector leftVector, MyVector rightVector) { return new MyVector(leftVector._x - rightVector._x, leftVector._y - rightVector._y); } public static MyVector operator *(MyVector aVector, double aScalar) { return new MyVector(aVector._x * aScalar, aVector._y * aScalar); } // Added using ReSharper due to CodeAnalysis nagging public bool Equals(MyVector other) { return _x.Equals(other._x) && _y.Equals(other._y); } public override bool Equals(object obj) { if (ReferenceEquals(null, obj)) return false; return obj is MyVector && Equals((MyVector) obj); } public override int GetHashCode() { unchecked { return (_x.GetHashCode()*397) ^ _y.GetHashCode(); } } public static bool operator ==(MyVector left, MyVector right) { return left.Equals(right); } public static bool operator !=(MyVector left, MyVector right) { return !left.Equals(right); } } 

And here's a little test program.

 public static class JustTesting { public static void Main() { Stopwatch stopwatch = new Stopwatch(); stopwatch.Start(); for (int i = 0; i < 10000000; i++) { TestIt(1, 0, 0, 0, 1, 1, 0.70710678118654757); TestIt(5, 4, 0, 0, 20, 10, 1.3416407864998738); TestIt(30, 15, 0, 0, 20, 10, 11.180339887498949); TestIt(-30, 15, 0, 0, 20, 10, 33.541019662496844); TestIt(5, 1, 0, 0, 10, 0, 1.0); TestIt(1, 5, 0, 0, 0, 10, 1.0); } stopwatch.Stop(); TimeSpan timeSpan = stopwatch.Elapsed; } private static void TestIt(float aPointX, float aPointY, float lineSegmentPoint1X, float lineSegmentPoint1Y, float lineSegmentPoint2X, float lineSegmentPoint2Y, double expectedAnswer) { // Katz double d1 = DistanceFromPointToLineSegment(new MyVector(aPointX, aPointY), new MyVector(lineSegmentPoint1X, lineSegmentPoint1Y), new MyVector(lineSegmentPoint2X, lineSegmentPoint2Y)); Debug.Assert(d1 == expectedAnswer); /* // Katz using squared distance double d2 = DistanceFromPointToLineSegmentSquared(new MyVector(aPointX, aPointY), new MyVector(lineSegmentPoint1X, lineSegmentPoint1Y), new MyVector(lineSegmentPoint2X, lineSegmentPoint2Y)); Debug.Assert(Math.Abs(d2 - expectedAnswer * expectedAnswer) < 1E-7f); */ /* // Matti (optimized) double d3 = FloatVector.DistanceToLineSegment(new PointF(aPointX, aPointY), new PointF(lineSegmentPoint1X, lineSegmentPoint1Y), new PointF(lineSegmentPoint2X, lineSegmentPoint2Y)); Debug.Assert(Math.Abs(d3 - expectedAnswer) < 1E-7f); */ } private static double DistanceFromPointToLineSegment(MyVector aPoint, MyVector lineSegmentPoint1, MyVector lineSegmentPoint2) { MyVector closestPoint; // Not used return aPoint.DistanceToLineSegment(lineSegmentPoint1, lineSegmentPoint2, out closestPoint); } private static double DistanceFromPointToLineSegmentSquared(MyVector aPoint, MyVector lineSegmentPoint1, MyVector lineSegmentPoint2) { MyVector closestPoint; // Not used return aPoint.DistanceToLineSegmentSquared(lineSegmentPoint1, lineSegmentPoint2, out closestPoint); } } 

As you can see, I tried to measure the difference between using the version that avoids the Sqrt() method and the normal version. My tests indicate you can maybe save about 2.5%, but I'm not even sure of that - the variations within the various test runs were of the same order of magnitude. I also tried measuring the version posted by Matti (plus an obvious optimization), and that version seems to be about 4% slower than the version based on Katz/Grumdrig code.

Edit: Incidentally, I've also tried measuring a method that finds the distance to an infinite line (not a line segment) using a cross product (and a Sqrt()), and it's about 32% faster.

Here is devnullicus's C++ version converted to C#. For my implementation I needed to know the point of intersection and found his solution to work well.

public static bool PointSegmentDistanceSquared(PointF point, PointF lineStart, PointF lineEnd, out double distance, out PointF intersectPoint) { const double kMinSegmentLenSquared = 0.00000001; // adjust to suit. If you use float, you'll probably want something like 0.000001f const double kEpsilon = 1.0E-14; // adjust to suit. If you use floats, you'll probably want something like 1E-7f double dX = lineEnd.X - lineStart.X; double dY = lineEnd.Y - lineStart.Y; double dp1X = point.X - lineStart.X; double dp1Y = point.Y - lineStart.Y; double segLenSquared = (dX * dX) + (dY * dY); double t = 0.0; if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared) { // segment is a point. intersectPoint = lineStart; t = 0.0; distance = ((dp1X * dp1X) + (dp1Y * dp1Y)); } else { // Project a line from p to the segment [p1,p2]. By considering the line // extending the segment, parameterized as p1 + (t * (p2 - p1)), // we find projection of point p onto the line. // It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2 t = ((dp1X * dX) + (dp1Y * dY)) / segLenSquared; if (t < kEpsilon) { // intersects at or to the "left" of first segment vertex (lineStart.X, lineStart.Y). If t is approximately 0.0, then // intersection is at p1. If t is less than that, then there is no intersection (i.e. p is not within // the 'bounds' of the segment) if (t > -kEpsilon) { // intersects at 1st segment vertex t = 0.0; } // set our 'intersection' point to p1. intersectPoint = lineStart; // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)). } else if (t > (1.0 - kEpsilon)) { // intersects at or to the "right" of second segment vertex (lineEnd.X, lineEnd.Y). If t is approximately 1.0, then // intersection is at p2. If t is greater than that, then there is no intersection (i.e. p is not within // the 'bounds' of the segment) if (t < (1.0 + kEpsilon)) { // intersects at 2nd segment vertex t = 1.0; } // set our 'intersection' point to p2. intersectPoint = lineEnd; // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)). } else { // The projection of the point to the point on the segment that is perpendicular succeeded and the point // is 'within' the bounds of the segment. Set the intersection point as that projected point. intersectPoint = new PointF((float)(lineStart.X + (t * dX)), (float)(lineStart.Y + (t * dY))); } // return the squared distance from p to the intersection point. Note that we return the squared distance // as an optimization because many times you just need to compare relative distances and the squared values // works fine for that. If you want the ACTUAL distance, just take the square root of this value. double dpqX = point.X - intersectPoint.X; double dpqY = point.Y - intersectPoint.Y; distance = ((dpqX * dpqX) + (dpqY * dpqY)); } return true; } 

A 2D and 3D solution

Consider a change of basis such that the line segment becomes (0, 0, 0)-(d, 0, 0) and the point (u, v, 0). The shortest distance occurs in that plane and is given by

 u ≤ 0 -> d(A, C) 0 ≤ u ≤ d -> |v| d ≤ u -> d(B, C) 

(the distance to one of the endpoints or to the supporting line, depending on the projection to the line. The iso-distance locus is made of two half-circles and two line segments.)

In the above expression, d is the length of the segment AB, and u, v are respectivey the scalar product and (modulus of the) cross product of AB/d (unit vector in the direction of AB) and AC. Hence vectorially,

AB.AC ≤ 0 -> |AC| 0 ≤ AB.AC ≤ AB² -> |ABxAC|/|AB| AB² ≤ AB.AC -> |BC| 

C# Version

public static FP DistanceToLineSegment(FPVector3 a, FPVector3 b, FPVector3 point) { var d = b - a; var s = d.SqrMagnitude; var ds = d / s; var lambda = FPVector3.Dot(point - a, ds); var p = FPMath.Clamp01(lambda) * d; return (a + p - point).Magnitude; } 

The above function is not working on vertical lines. Here is a function that is working fine! Line with points p1, p2. and CheckPoint is p;

public float DistanceOfPointToLine2(PointF p1, PointF p2, PointF p) { // (y1-y2)x + (x2-x1)y + (x1y2-x2y1) //d(P,L) = -------------------------------- // sqrt( (x2-x1)pow2 + (y2-y1)pow2 ) double ch = (p1.Y - p2.Y) * p.X + (p2.X - p1.X) * p.Y + (p1.X * p2.Y - p2.X * p1.Y); double del = Math.Sqrt(Math.Pow(p2.X - p1.X, 2) + Math.Pow(p2.Y - p1.Y, 2)); double d = ch / del; return (float)d; } 

see the Matlab GEOMETRY toolbox in the following website: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html

ctrl+f and type "segment" to find line segment related functions. the functions "segment_point_dist_2d.m" and "segment_point_dist_3d.m" are what you need.

The GEOMETRY codes are available in a C version and a C++ version and a FORTRAN77 version and a FORTRAN90 version and a MATLAB version.

AutoHotkeys version based on Joshua's Javascript:

plDist(x, y, x1, y1, x2, y2) { A:= x - x1 B:= y - y1 C:= x2 - x1 D:= y2 - y1 dot:= A*C + B*D sqLen:= C*C + D*D param:= dot / sqLen if (param < 0 || ((x1 = x2) && (y1 = y2))) { xx:= x1 yy:= y1 } else if (param > 1) { xx:= x2 yy:= y2 } else { xx:= x1 + param*C yy:= y1 + param*D } dx:= x - xx dy:= y - yy return sqrt(dx*dx + dy*dy) } 

the accepted answer does not work (e.g. distance between 0,0 and (-10,2,10,2) should be 2).

here's code that works:

 def dist2line2(x,y,line): x1,y1,x2,y2=line vx = x1 - x vy = y1 - y ux = x2-x1 uy = y2-y1 length = ux * ux + uy * uy det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment if det < 0: return (x1 - x)**2 + (y1 - y)**2 if det > length: return (x2 - x)**2 + (y2 - y)**2 det = ux * vy - uy * vx return det**2 / length def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line)) 

Didn't see a Java implementation here, so I translated the Javascript function from the accepted answer to Java code:

static double sqr(double x) { return x * x; } static double dist2(DoublePoint v, DoublePoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); } static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) { double l2 = dist2(v, w); if (l2 == 0) return dist2(p, v); double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2; if (t < 0) return dist2(p, v); if (t > 1) return dist2(p, w); return dist2(p, new DoublePoint( v.x + t * (w.x - v.x), v.y + t * (w.y - v.y) )); } static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) { return Math.sqrt(distToSegmentSquared(p, v, w)); } static class DoublePoint { public double x; public double y; public DoublePoint(double x, double y) { this.x = x; this.y = y; } } 

WPF version:

public class LineSegment { private readonly Vector _offset; private readonly Vector _vector; public LineSegment(Point start, Point end) { _offset = (Vector)start; _vector = (Vector)(end - _offset); } public double DistanceTo(Point pt) { var v = (Vector)pt - _offset; // first, find a projection point on the segment in parametric form (0..1) var p = (v * _vector) / _vector.LengthSquared; // and limit it so it lays inside the segment p = Math.Min(Math.Max(p, 0), 1); // now, find the distance from that point to our point return (_vector * p - v).Length; } }