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;

我需要一个Godot (GDscript)的实现，所以我写了一个基于grumdrig接受的答案:

func minimum_distance(v: Vector2, w: Vector2, p: Vector2):
    # Return minimum distance between line segment vw and point p
    var l2: float = (v - w).length_squared()  # i.e. |w-v|^2 -  avoid a sqrt
    if l2 == 0.0:
        return p.distance_to(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.
    var t: float = max(0, min(1, (p - v).dot(w - v) / l2))
    var projection: Vector2 = v + t * (w - v)  # Projection falls on the segment
    
    return p.distance_to(projection)

在f#中，点c到a和b之间的线段的距离为:

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

向量d沿着线段从a指向b。d/s与c-a的点积给出了无限直线与点c之间最接近点的参数。使用min和max函数将该参数钳制到范围0..s，使该点位于a和b之间。最后，a+p-c的长度是c到线段上最近点的距离。

使用示例:

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

这是一个自成体系的Delphi / Pascal版本的函数，基于上面约书亚的答案。使用TPoint用于VCL屏幕图形，但应该易于根据需要进行调整。

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 <> 0) then
   begin
      param := ((a * c) + (b * d)) / len_sq;
   end;

   if param < 0 then
   begin
      xx := pt1.x;
      yy := pt1.y;
   end
   else if param > 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;

使用arctangents的一行解决方案:

思路是将A移动到(0,0)，并顺时针旋转三角形，使C位于X轴上， 当这种情况发生时，By就是距离。

a角= Atan(Cy - Ay, Cx - Ax); b角= Atan(By - Ay, Bx - Ax); AB长度=平方根((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;
}

一行c#(要转换为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))

特征c++版本的3D线段和点

// Return minimum distance between line segment: head--->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--->tail.

    double t = max(0,min(1,(point-head).dot(tail-head)/l2));
    Eigen::Vector3d projection = head + t*(tail-head);

    return (point - projection).norm();
}