本想在GLSL中这样做，但如果可能的话，最好避免所有这些条件。使用clamp()可以避免两种端点情况:

// 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 <= 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
}

如果您可以确定A和B彼此不会非常接近，则可以简化为删除If()。事实上，即使A和B是相同的，我的GPU仍然给出了这个无条件版本的正确结果(但这是使用pre-OpenGL 4.1，其中GLSL除零是未定义的):

// 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);
}

计算距离是很简单的——GLSL提供了一个distance()函数，你可以在这个最近的点和P。

灵感来自Iñigo Quilez的胶囊距离函数代码

这是Javascript中最简单的完整代码。

(X, y)是目标点(x1, y)到(x2, y)是线段。

更新:修复了评论中0长度的行问题。

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);
}

更新:Kotlin版本

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)
}

Grumdrig的c++ /JavaScript实现对我来说非常有用，所以我提供了我正在使用的Python直接端口。完整的代码在这里。

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)

在数学

它使用线段的参数描述，并将点投影到线段定义的直线中。当参数在线段内从0到1时，如果投影在这个范围之外，我们计算到相应端点的距离，而不是法线到线段的直线。

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*)
    ]
   ];  

策划的结果:

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

画出比截断距离更近的点:

等高线图:

在我自己的问题线程如何计算在C, c# / .NET 2.0或Java的所有情况下一个点和线段之间的最短2D距离?当我找到一个c#的答案时，我被要求把它放在这里:所以它是从http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static修改的:

//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);
} 

我不是要回答问题，而是要问问题，所以我希望我不会因为某些原因而得到数百万张反对票，而是批评。我只是想(并被鼓励)分享其他人的想法，因为这个帖子中的解决方案要么是用一些奇异的语言(Fortran, Mathematica)，要么被某人标记为错误。对我来说唯一有用的(由Grumdrig编写)是用c++编写的，没有人标记它有错误。但是它缺少被调用的方法(dot等)。

这里是与c++答案相同的东西，但移植到pascal。点参数的顺序已经改变，以适应我的代码，但还是一样的东西。

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 < 0.0) then begin
    result := Distance(p, v);       // Beyond the 'v' end of the segment
    exit;
  end
  else if (t > 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;