现在我的解决方案...... (Javascript)

这是非常快的，因为我试图避免任何数学。战俘的功能。

如你所见，在函数的最后，我得到了直线的距离。

代码来自lib http://www.draw2d.org/graphiti/jsdoc/#!/例子

/**
 * 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;
};

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;

对于感兴趣的人，这里是Joshua的Javascript代码到Objective-C的简单转换:

- (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);
}

我需要这个解决方案与MKMapPoint一起工作，所以我将分享它，以防其他人需要它。只是一些小的改变，这将返回米为单位的距离:

- (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));
}

这是一个基于向量数学的;这个解决方案也适用于更高的维度，并报告交点(在线段上)。

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)

结果是

2.1213203435596424
[[2.5]
 [2.5]]

这里解释了数学

https://brilliant.org/wiki/distance-between-point-and-line/

上面的函数在垂直线上不起作用。这是一个工作正常的函数! 与点p1 p2相交。CheckPoint为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;
}

用t-sql编码

点为(@px， @py)，线段从(@ax， @ay)到(@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