Shortest distance between a point and a line segment

Question

I need a basic function to find the shortest distance between a point and a line segment   Feel free to write the solution in any language you want  I can translate it into what I m using  Javascript    EDIT  My line segment is defined by two endpoints  So my line segment AB is defined by the two points A  x1 y1  and B  x2 y2    I m trying to find the distance between this line segment and a point C  x3 y3    My geometry skills are rusty  so the examples I ve seen are confusing  I m sorry to admit

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Python Numpy implementation for 2D coordinate array   import numpy as np   def dist2d p1  p2  coords           Distance from points to a finite line btwn p1 - gt  p2         assert coords ndim    2 and coords shape 1     2   coords is not 2 dim      dp   p2 - p1     st   dp 0   2   dp 1   2     u     coords    0  - p1 0     dp 0     coords    1  - p1 1     dp 1     st      u u  gt  1     1      u u  lt  0     0       dx    p1 0    u   dp 0   - coords    0      dy    p1 1    u   dp 1   - coords    1       return np sqrt dx  2   dy  2      Usage  p1   np array  0   0    p2   np array  0   10       List of coordinates coords   np array        0   0          5   5          10   10          20   20            d   dist2d p1  p2  coords     Single coordinate coord   np array  25   25    d   dist2d p1  p2  coord np newaxis

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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  lt  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  lt   0              closestPoint   lineSegmentPoint1           else if  t  gt   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   amp  amp   y Equals other  y                  public override bool Equals object obj                   if  ReferenceEquals null  obj   return false           return obj is MyVector  amp  amp  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  lt  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   lt  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   lt  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

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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  lt  0 0f  return dist2 p  v       if  t  gt  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

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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

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Here is one based on vector math  this solution will also work for higher dimensions and also report on the interection point  on the line segment    def dist x1 y1 x2 y2 px py       a   np array   x1 y1    T     b   np array   x2 y2    T     x   np array   px py    T     tp    np dot x T  b  - np dot a T  b     np dot b T  b      tp   tp 0  0      tmp   x -  a   tp b      d   np sqrt np dot tmp T tmp  0  0       return d  a tp b  x1 y1 2  2  x2 y2 5  5  px py 4  1   d  inters   dist x1 y1  x2 y2  px py  print  d  print  inters    Result is  2 1213203435596424   2 5    2 5     The math is explained here  https   brilliant org wiki distance-between-point-and-line

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Solution for dart and flutter   import  dart math  as math   class Utils      static double shortestDistance Point p1  Point p2  Point p3         double px   p2 x - p1 x        double py   p2 y - p1 y        double temp    px px     py py         double u     p3 x - p1 x  px    p3 y - p1 y   py   temp        if u gt 1           u 1                else if u lt 0           u 0                double x   p1 x   u px        double y   p1 y   u py        double dx   x - p3 x        double dy   y - p3 y        double dist   math sqrt dx dx dy dy         return dist          class Point     double x    double y    Point this x  this y

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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  gt  1          u   1     elif u  lt  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  gt  1          u   1      else if  u  lt  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 gt 1               u 1                    else if u lt 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

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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  lt  0     l1 x    l2 x  amp  amp  l1 y    l2 y             xx   l1 x          yy   l1 y            else if  param  gt  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  lt  0     l1 x    l2 x  amp  amp  l1 y    l2 y             xx   l1 x          yy   l1 y            else if  param  gt  1            xx   l2 x          yy   l2 y            else           xx   l1 x   param   C          yy   l1 y   param   D             return MKMetersBetweenMapPoints p  MKMapPointMake xx  yy

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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  lt  0  return dist2 p  v       if  t  gt  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

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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

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Lua  Finds minimum distance between a line segment not the whole line  and a point   function solveLinearEquation A1 B1 C1 A2 B2 C2  --it is the implitaion of a method of solving linear equations in x and y   local f1   B1 C2 -B2 C1   local f2   A2 C1-A1 C2   local f3   A1 B2 -A2 B1   return  x  f1 f3  y  f2 f3  end   function pointLiesOnLine x y x1 y1 x2 y2    local dx1   x-x1   local  dy1   y-y1   local dx2   x-x2   local  dy2   y-y2   local crossProduct   dy1 dx2 -dx1 dy2  if crossProduct    0  then  return  false else   if   x1 gt  x  and  x gt  x2   or   x2 gt  x  and  x gt  x1   then     if   y1 gt  y  and  y gt  y2   or   y2 gt  y  and  y gt  y1   then       return true     else return false end   else  return false end end end   function dist x1 y1 x2 y2    local dx   x1-x2   local dy   y1-y2   return math sqrt dx dx   dy  dy   end   function findMinDistBetnPointAndLine x1 y1 x2 y2 x3 y3  -- finds the min  distance between  x3 y3  and line  x1 y2 -- x2 y2     local A2 B2 C2 A1 B1 C1    local dx   y2-y1    local dy   x2-x1    if dx    0 then A2 1 B2 0 C2 -x3 A1 0 B1 1 C1 -y1     elseif dy    0 then A2 0 B2 1 C2 -y3 A1 1 B1 0 C1 -x1    else       local m1   dy dx       local m2   -1 m1       A2 m2 B2 -1 C2 y3-m2 x3 A1 m1 B1 -1 C1 y1-m1 x1    end  local intsecPoint  solveLinearEquation A1 B1 C1 A2 B2 C2  if pointLiesOnLine intsecPoint x  intsecPoint y x1 y1 x2 y2  then    return dist intsecPoint x  intsecPoint y  x3 y3   else    return math min dist x3 y3 x1 y1  dist x3 y3 x2 y2   end end

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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 amp d2 geometry1 amp 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  gt  0               return Distance pointB  pointC            double dot2   DotProduct pointB  pointA  pointC           if  dot2  gt  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

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Eli  the code you ve settled on is incorrect  A point near the line on which the segment lies but far off one end of the segment would be incorrectly judged near the segment  Update  The incorrect answer mentioned is no longer the accepted one   Here s some correct code  in C    It presumes a class 2D-vector class vec2  float x y    essentially  with operators to add  subract  scale  etc  and a distance and dot product function  i e  x1 x2   y1 y2    float minimum distance vec2 v  vec2 w  vec2 p         Return minimum distance between line segment vw and point p   const float l2   length squared v  w       i e   w-v  2 -  avoid a sqrt   if  l2    0 0  return distance p  v        v    w case      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      We clamp t from  0 1  to handle points outside the segment vw    const float t   max 0  min 1  dot p - v  w - v    l2      const vec2 projection   v   t    w - v       Projection falls on the segment   return distance p  projection       EDIT  I needed a Javascript implementation  so here it is  with no dependencies  or comments  but it s a direct port of the above   Points are represented as objects with x and y attributes   function sqr x    return x   x   function dist2 v  w    return sqr v x - w x    sqr v y - w y    function distToSegmentSquared p  v  w      var l2   dist2 v  w     if  l2    0  return dist2 p  v     var t     p x - v x     w x - v x     p y - v y     w y - v y     l2    t   Math max 0  Math min 1  t      return dist2 p    x  v x   t    w x - v x                       y  v y   t    w y - v y        function distToSegment p  v  w    return Math sqrt distToSegmentSquared p  v  w        EDIT 2  I needed a Java version  but more important  I needed it in 3d instead of 2d   float dist to segment squared float px  float py  float pz  float lx1  float ly1  float lz1  float lx2  float ly2  float lz2      float line dist   dist sq lx1  ly1  lz1  lx2  ly2  lz2     if  line dist    0  return dist sq px  py  pz  lx1  ly1  lz1     float t     px - lx1     lx2 - lx1     py - ly1     ly2 - ly1     pz - lz1     lz2 - lz1     line dist    t   constrain t  0  1     return dist sq px  py  pz  lx1   t    lx2 - lx1   ly1   t    ly2 - ly1   lz1   t    lz2 - lz1

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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

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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 - gt  d A  C  0   u   d - gt   v  d   u     - gt  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             - gt   AC      0   AB AC   AB     - gt   ABxAC   AB            AB     AB AC - gt   BC

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Matlab direct Grumdrig implementation  function ans distP2S px py vx vy wx wy     px py vx vy wx wy    t    px-vx   wx-vx   py-vy   wy-vy    idist vx wx vy wy  2     idist px vx py vy  idist px vx t  wx-vx  py vy t  wy-vy   idist px wx py wy       ans 1  t gt 0   t gt 1       lt 0 0 lt  t lt  1 t gt 1       end  function d idist a b c d   d abs a-b 1i  c-d    end

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Swift implementation of http   paulbourke net geometry pointlineplane source c      static func magnitude p1  CGPoint  p2  CGPoint  - gt  CGFloat           let vector   CGPoint x  p2 x - p1 x  y  p2 y - p1 y          return sqrt pow vector x  2    pow vector y  2                  http   paulbourke net geometry pointlineplane          http   paulbourke net geometry pointlineplane source c     static func pointDistanceToLine point  CGPoint  lineStart  CGPoint  lineEnd  CGPoint  - gt  CGFloat             let lineMag   magnitude p1  lineEnd  p2  lineStart          let u      point x - lineStart x     lineEnd x - lineStart x                       point y - lineStart y     lineEnd y - lineStart y                       lineMag   lineMag           if u  lt  0    u  gt  1                  closest point does not fall within the line segment             return nil                    let intersectionX   lineStart x   u    lineEnd x - lineStart x          let intersectionY   lineStart y   u    lineEnd y - lineStart y           return magnitude p1  point  p2  CGPoint x  intersectionX  y  intersectionY

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If its an infinite line  not a line segment  the simplest way is this  in ruby   where mx   b is the line and  x1  y1  is the known point   y1 - mx1 - b  abs   Math sqrt m  2   1

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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

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A little cleaner solution in JavaScript based on this formula    distToSegment  function  point  linePointA  linePointB        var x0   point X      var y0   point Y       var x1   linePointA X      var y1   linePointA Y       var x2   linePointB X      var y2   linePointB Y       var Dx    x2 - x1       var Dy    y2 - y1        var numerator   Math abs Dy x0 - Dx y0 - x1 y2   x2 y1       var denominator   Math sqrt Dx Dx   Dy Dy       if  denominator    0            return this dist2 point  linePointA              return numerator denominator

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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  lt  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  lt  0           r   norm xy0-xyP 2       elseif  detP  gt  lenSqr           r   norm xy1-xyP 2       else         r   abs ux vy-uy vx  sqrt lenSqr       end end       function selfTest             ok lt  NASGU gt          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

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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  gt   -kMinSegmentLenSquared  amp  amp  segLenSquared  lt   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  lt  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  gt  -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  gt   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  lt   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

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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  lt   0      pt is  behind  st  use st               nearestPt   st             else if  keyDot  gt   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

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Matlab solution by Tim from Cody function ans distP2S x0 y0 x1 y1 x2 y2    Point is x0 y0 z complex x0-x1 y0-y1   complex x2-x1 y2-y1   abs z-ans min 1 max 0 real z ans

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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

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Eigen C   Version for 3D line segment and point     Return minimum distance between line segment  head--- gt tail and point double MinimumDistance Eigen  Vector3d head  Eigen  Vector3d tail Eigen  Vector3d point        double l2   std  pow  head - tail  norm   2       if l2   0 0  return  head - point  norm      head    tail case         Consider the line extending the segment  parameterized as head   t  tail - point          We find projection of point onto the line         It falls where t     point-head     tail-head      tail-head  2        We clamp t from  0 1  to handle points outside the segment head--- gt tail       double t   max 0 min 1  point-head  dot tail-head  l2        Eigen  Vector3d projection   head   t  tail-head        return  point - projection  norm

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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  lt  0           return dbo fn Distance  px   py   ax   ay       if   t  gt  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

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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  lt  0        xx   x1      yy   y1        else if  param  gt  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

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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  lt  0  EuclideanDistance start  pt                     If outside bounds       param  gt  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

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Just came across this and thought I d add a Lua implementation  It assumes that points are given as tables  x xVal  y yVal  and the line or segment is given by a table containing two points  see example below    function distance  P1  P2       return math sqrt  P1 x-P2 x  2    P1 y-P2 y  2  end  -- Returns false if the point lies beyond the reaches of the segment function distPointToSegment  line  P       if line 1  x    line 2  x and line 1  y    line 2  y then         print  Error  Not a line            return false     end      local d   distance  line 1   line 2         local t     P x - line 1  x   line 2  x - line 1  x     P y - line 1  y   line 2  y - line 1  y    d 2       local projection          projection x   line 1  x   t  line 2  x-line 1  x      projection y   line 1  y   t  line 2  y-line 1  y       if t  gt   0 and t  lt   1 then   -- within line segment          return distance  projection   x P x  y P y        else         return false     end end  -- Returns value even if point is further down the line  outside segment  function distPointToLine  line  P       if line 1  x    line 2  x and line 1  y    line 2  y then         print  Error  Not a line            return false     end      local d   distance  line 1   line 2         local t     P x - line 1  x   line 2  x - line 1  x     P y - line 1  y   line 2  y - line 1  y    d 2       local projection          projection x   line 1  x   t  line 2  x-line 1  x      projection y   line 1  y   t  line 2  y-line 1  y       return distance  projection   x P x  y P y    end   Example usage   local P1    x   0  y   0  local P2    x   10  y   10  local line     P1  P2   local P3    x   7  y   15  print distPointToLine  line  P3     -- prints 5 6568542494924 print distPointToSegment  line  P3    -- prints false

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Here s the code I ended up writing   This code assumes that a point is defined in the form of  x 5  y 7    Note that this is not the absolute most efficient way  but it s the simplest and easiest-to-understand code that I could come up with      a  b  and c in the code below are all points  function distance a  b        var dx   a x - b x      var dy   a y - b y      return Math sqrt dx dx   dy dy      function Segment a  b        var ab             x  b x - a x          y  b y - a y            var length   distance a  b        function cross c            return ab x    c y-a y  - ab y    c x-a x               this distanceFrom   function c            return Math min distance a c                           distance b c                           Math abs cross c    length

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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  lt  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 amp  t                                      double amp  qx  double amp  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  gt   -kMinSegmentLenSquared  amp  amp  segLenSquared  lt   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  lt  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  gt  -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  gt   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  lt   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

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Here is SQL implementation for HSQLDB  CREATE FUNCTION dist to segment px double  py double  vx double  vy double  wx double  wy double    RETURNS double BEGIN atomic    declare l2 double     declare t double     declare nx double     declare ny double     set l2   vx - wx   vx - wx     vy - wy   vy - wy      IF l2   0 THEN      RETURN sqrt  vx - px   vx - px     vy - py   vy - py       ELSE      set t     px - vx     wx - vx     py - vy     wy - vy     l2       set t   GREATEST 0  LEAST 1  t         set nx vx   t    wx - vx        set ny vy   t    wy - vy        RETURN sqrt  nx - px   nx - px     ny - py   ny - py       END IF  END   And implementation for Postgres  CREATE FUNCTION dist to segment px numeric  py numeric  vx numeric  vy numeric  wx numeric  wy numeric    RETURNS numeric AS       declare l2 numeric     declare t numeric     declare nx numeric     declare ny numeric  BEGIN     l2     vx - wx   vx - wx     vy - wy   vy - wy      IF l2   0 THEN      RETURN sqrt  vx - px   vx - px     vy - py   vy - py       ELSE      t      px - vx     wx - vx     py - vy     wy - vy     l2       t    GREATEST 0  LEAST 1  t         nx    vx   t    wx - vx        ny    vy   t    wy - vy        RETURN sqrt  nx - px   nx - px     ny - py   ny - py       END IF  END     LANGUAGE plpgsql

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Same as this answer  except in Visual Basic  Makes it usable as a User Defined Function in Microsoft Excel and VBA macros   The function returns the closest distance from point  x y  to the segment defined by  x1 y1  and  x2 y2   Function DistanceToSegment x As Double  y As Double  x1 As Double  y1 As Double  x2 As Double  y2 As Double     Dim A As Double   A   x - x1   Dim B As Double   B   y - y1   Dim C  As Double   C   x2 - x1   Dim D As Double   D   y2 - y1    Dim dot As Double   dot   A   C   B   D   Dim len sq As Double   len sq   C   C   D   D   Dim param As Double   param   -1    If  len sq  lt  gt  0  Then       param   dot   len sq   End If    Dim xx As Double   Dim yy As Double    If  param  lt  0  Then     xx   x1     yy   y1   ElseIf  param  gt  1  Then     xx   x2     yy   y2   Else     xx   x1   param   C     yy   y1   param   D   End If    Dim dx As Double   dx   x - xx   Dim dy As Double   dy   y - yy    DistanceToSegment   Math Sqr dx   dx   dy   dy   End Function

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Here is same thing as the C   answer but ported to pascal  The order of the point parameter has changed to suit my code but is the same thing   function Dot const p1  p2  PointF   double  begin   Result    p1 x   p2 x   p1 y   p2 y  end  function SubPoint const p1  p2  PointF   PointF  begin   result x    p1 x - p2 x    result y    p1 y - p2 y  end   function ShortestDistance2 const p v w   PointF    double  var   l2 t   double    projection tt  PointF  begin      Return minimum distance between line segment vw and point p     l2    length squared v  w       i e   w-v  2 -  avoid a sqrt   l2    Distance v w     l2    MPower l2 2     if  l2   0 0  then begin     result   Distance p  v        v    w case     exit    end       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    Dot SubPoint p v  SubPoint w v     l2    if  t  lt  0 0  then begin     result    Distance p  v            Beyond the  v  end of the segment     exit    end   else if  t  gt  1 0  then begin     result    Distance p  w       Beyond the  w  end of the segment     exit    end      projection    v   t    w - v       Projection falls on the segment   tt x    v x   t    w x - v x     tt y    v y   t    w y - v y     result    Distance p  tt   end

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In javascript using Geometry   var a     x 20  y 20    start segment     var b     x 40  y 30    end segment    var c     x 37  y 14    point        magnitude from a to c     var ac   Math sqrt  Math pow    a x - c x    2     Math pow    a y - c y    2            magnitude from b to c    var bc   Math sqrt  Math pow    b x - c x    2     Math pow    b y - c y    2             magnitude from a to b  base       var ab   Math sqrt  Math pow    a x - b x    2     Math pow    a y - b y    2              perimeter of triangle      var p   ac   bc   ab          area of the triangle     var area   Math sqrt  p 2     p 2 - ac      p 2 - bc       p 2 - ab              height of the triangle   distance     var h     area   2     ab      console log   height      h

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This answer is based on the accepted answer s JavaScript solution  It s mainly just formatted nicer  with longer function names  and of course shorter function syntax because it s in ES6   CoffeeScript   JavaScript version  ES6   distanceSquared    v  w   gt  Math pow v x - w x  2    Math pow v y - w y  2   distance    v  w   gt  Math sqrt distanceSquared v  w     distanceToLineSegmentSquared    p  v  w   gt        l2   distanceSquared v  w       if  l2     0            return distanceSquared p  v             t     p x - v x     w x - v x     p y - v y     w y - v y     l2      t   Math max 0  Math min 1  t        return distanceSquared p            x  v x   t    w x - v x           y  v y   t    w y - v y            distanceToLineSegment    p  v  w   gt        return Math sqrt distanceToLineSegmentSquared p  v        CoffeeScript version  distanceSquared    v  w - gt   v x - w x     2    v y - w y     2 distance    v  w - gt  Math sqrt distanceSquared v  w    distanceToLineSegmentSquared    p  v  w - gt      l2   distanceSquared v  w      return distanceSquared p  v  if l2 is 0     t     p x - v x     w x - v x     p y - v y     w y - v y     l2     t   Math max 0  Math min 1  t       distanceSquared p            x  v x   t    w x - v x          y  v y   t    w y - v y          distanceToLineSegment    p  v  w - gt      Math sqrt distanceToLineSegmentSquared p  v  w

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Here it is using Swift         Distance from a point  p1  to line l1 l2    func distanceFromPoint p  CGPoint  toLineSegment l1  CGPoint  and l2  CGPoint  - gt  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     let param   dot   len sq      var xx  yy  CGFloat      if param  lt  0     l1 x    l2 x  amp  amp  l1 y    l2 y            xx   l1 x         yy   l1 y       else if param  gt  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

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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   gt  max 0 0   gt  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

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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  lt  0      x1   x2   amp  amp   y1   y2              xx   x1         yy   y1       else if  param  gt  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

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Wanted to do this in GLSL  but it s better to avoid all those conditionals if possible  Using clamp   avoids the two end-point cases      find closest point to P on line segment AB  vec3 closest point on line segment in vec3 P  in vec3 A  in vec3 B        vec3 AP   P - A  AB   B - A      float l   dot AB  AB       if  l  lt   0 0000001  return A        A and B are practically the same     return AP - AB clamp dot AP  AB  l  0 0  1 0       do the projection     If you can be sure that A and B are never very close to each other  this can be simplified to remove the if    In fact  even if A and B are the same  my GPU still gives the right result with this condition-free version  but this is using pre-OpenGL 4 1 in which GLSL divide by zero is undefined       find closest point to P on line segment AB  vec3 closest point on line segment in vec3 P  in vec3 A  in vec3 B        vec3 AP   P - A  AB   B - A      return AP - AB clamp dot AP  AB  dot AB  AB   0 0  1 0       To compute the distance is trivial -- GLSL provides a distance   function which you can use on this closest point and P   Inspired by I  igo Quilez s code for a capsule distance function

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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  lt   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  gt X2 Y2          to get px py to be the negative of px py  gt 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  lt   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  gt X1 Y1 vector times the              length of the X2 Y2  gt 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  lt  0            lenSq   0            return Math sqrt lenSq  lt coronaWidth

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I ve made an interactive Desmos graph to demonstrate how to achieve this   https   www desmos com calculator kswrm8ddum  The red point is A  the green point is B  and the point C is blue  You can drag the points in the graph to see the values change  On the left  the value  s  is the parameter of the line segment  i e  s   0 means the point A  and s   1 means the point B   The value  d  is the distance from the third point to the line through A and B   EDIT   Fun little insight  the coordinate  s  d  is the coordinate of the third point C in the coordinate system where AB is the unit x-axis  and the unit y-axis is perpendicular to AB

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Here s a self contained Delphi   Pascal version of the function based on Joshua s answer up above  Uses TPoint for VCL screen graphics  but should be easy to adjust as needed   function DistancePtToSegment  pt  pt1  pt2  TPoint   double  var    a  b  c  d  double     len sq  double     param  double     xx  yy  double     dx  dy  double  begin    a    pt x - pt1 x     b    pt y - pt1 y     c    pt2 x - pt1 x     d    pt2 y - pt1 y      len sq     c   c     d   d      param    -1      if  len sq  lt  gt  0  then    begin       param      a   c     b   d     len sq     end      if param  lt  0 then    begin       xx    pt1 x        yy    pt1 y     end    else if param  gt  1 then    begin       xx    pt2 x        yy    pt2 y     end    else begin       xx    pt1 x   param   c        yy    pt1 y   param   d     end      dx    pt x - xx     dy    pt y - yy     result    sqrt   dx   dx     dy   dy   end

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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  amp v   const               return Vec2  this- gt  x   v  x  this- gt  y   v  y               Vec2 operator-  const Vec2  amp v   const               return Vec2  this- gt  x - v  x  this- gt  y - v  y               Vec2 operator   const float f   const               return Vec2  this- gt  x   f  this- gt  y   f               float DistanceToSquared  const Vec2 p   const               const float dX   p  x - this- gt  x          const float dY   p  y - this- gt  y           return dX   dX   dY   dY             float DistanceTo  const Vec2 p   const               return sqrt  this- gt DistanceToSquared  p                 float DotProduct  const Vec2 p   const               return this- gt  x   p  x   this- gt  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  lt  0 0                    beyond the v end of the segment           q    v           return v DistanceTo  p              else if   t  gt  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     amp 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   amp qX   amp 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

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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  lt  0        Beyond v end of the segment     return distance squared p  v    elif t  gt  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

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in R       distance beetween segment ab and point c in 2D space getDistance ort 2  lt - function a  b  c      go to complex numbers   A lt -c a 1  1i a 2  b 1  1i b 2     q c 1  1i c 2        function to get coefficients of line  ab    getAlphaBeta  lt - function A      a lt -Re A 2  -Re A 1       b lt -Im A 2  -Im A 1       ab lt -as numeric       ab 1   lt - -Re A 1   b a Im A 1       ab 2   lt -b a     if Im A 1    Im A 2    ab lt - c Im A 1   0      if Re A 1    Re A 2    ab  lt - NA     return ab            function to get coefficients of line ortogonal to line  ab  which goes through point q   getAlphaBeta ort lt -function A q      ab  lt - getAlphaBeta A     coef lt -c Re q  ab 2  Im q  -1 ab 2     if Re A 1    Re A 2    coef lt -c Im q  0    return coef            function to get coordinates of interception point     between line  ab  and its ortogonal which goes through point q   getIntersection ort  lt - function A  q       A ab  lt - getAlphaBeta A      q ab  lt - getAlphaBeta ort A q      if   is na A ab 1   amp A ab 2   0          x lt -Re q        y lt -Im A 1        if  is na A ab 1            x lt -Re A 1         y lt -Im q             if   is na A ab 1   amp A ab 2   0          x  lt -  q ab 1  - A ab 1    A ab 2  - q ab 2         y  lt - q ab 1    q ab 2  x      xy  lt - x   1i y       return xy           intersect lt -getIntersection ort A q    if   Mod A 1 -intersect  Mod A 2 -intersect   gt Mod A 1 -A 2     dist lt -min Mod A 1 -q  Mod A 2 -q       else dist lt -Mod q-intersect    return dist

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GLSL version      line  a - gt  b   point p enter image description here  1  float distanceToLine vec2 a  vec2 b  vec2 p        float aside   dot  p - a   b - a        if aside lt  0 0  return length p-a       float bside   dot  p - b   a - b        if bside lt  0 0  return length p-b       vec2 pointOnLine    bside a   aside b  pow length a-b  2 0       return length p - pointOnLine

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This algorithm is based on finding the intersection between the specified line and the orthogonal line  which contains the specified point  and calculating its distance  In case of a line segment  we must check if the intersection is between points of the line segment  if that s not the case then the minimum distance is between the specified point and one of the ending points of the line segment  This is a C  implementation   Double Distance Point a  Point b        double xdiff   a X - b X  ydiff   a Y - b Y      return Math Sqrt  long xdiff   xdiff    long ydiff   ydiff      Boolean IsBetween double x  double a  double b        return   a  lt   b  amp  amp  x  gt   a  amp  amp  x  lt   b      a  gt  b  amp  amp  x  lt   a  amp  amp  x  gt   b       Double GetDistance Point pt  Point pt1  Point pt2  out Point intersection        Double a  x  y  R       if  pt1 X    pt2 X            a    double  pt2 Y - pt1 Y     pt2 X - pt1 X           x    a    pt Y - pt1 Y    a   a   pt1 X   pt X     a   a   1           y   a   x   pt1 Y - a   pt1 X        else   x   pt1 X   y   pt Y         if  IsBetween x  pt1 X  pt2 X   amp  amp  IsBetween y  pt1 Y  pt2 Y             intersection   new Point  int x   int y           R   Distance intersection  pt         else           double d1   Distance pt  pt1   d2   Distance pt  pt2           if  d1  lt  d2    intersection   pt1  R   d1            else   intersection   pt2  R   d2          return R

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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  lt  0 or  gt  length then its outside the line segment      if det  lt  0         return  x1 - x   2    y1 - y   2      if det  gt  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

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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

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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      a angle   Atan Cy - Ay  Cx - Ax   b angle   Atan By - Ay  Bx - Ax   AB length   Sqrt   Bx - Ax  2     By - Ay  2   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

[language-agnostic] Shortest distance between a point and a line segment

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