我制作了一个交互式Desmos图来演示如何实现这一点:

https://www.desmos.com/calculator/kswrm8ddum

红点是A点，绿点是B点，C点是蓝色点。 您可以拖动图形中的点来查看值的变化。 左边的值“s”是线段的参数(即s = 0表示点A, s = 1表示点B)。 值“d”是第三点到经过A和B的直线的距离。

编辑:

有趣的小见解:坐标(s, d)是坐标系中第三点C的坐标，AB是单位x轴，单位y轴垂直于AB。

这是一个为有限线段而做的实现，而不是像这里的大多数其他函数那样的无限线(这就是为什么我做这个)。

Paul Bourke的理论实施。

Python:

def dist(x1, y1, x2, y2, x3, y3): # x3,y3 is the point
    px = x2-x1
    py = y2-y1

    norm = px*px + py*py

    u =  ((x3 - x1) * px + (y3 - y1) * py) / float(norm)

    if u > 1:
        u = 1
    elif u < 0:
        u = 0

    x = x1 + u * px
    y = y1 + u * py

    dx = x - x3
    dy = y - y3

    # Note: If the actual distance does not matter,
    # if you only want to compare what this function
    # returns to other results of this function, you
    # can just return the squared distance instead
    # (i.e. remove the sqrt) to gain a little performance

    dist = (dx*dx + dy*dy)**.5

    return dist

AS3:

public static function segmentDistToPoint(segA:Point, segB:Point, p:Point):Number
{
    var p2:Point = new Point(segB.x - segA.x, segB.y - segA.y);
    var something:Number = p2.x*p2.x + p2.y*p2.y;
    var u:Number = ((p.x - segA.x) * p2.x + (p.y - segA.y) * p2.y) / something;

    if (u > 1)
        u = 1;
    else if (u < 0)
        u = 0;

    var x:Number = segA.x + u * p2.x;
    var y:Number = segA.y + u * p2.y;

    var dx:Number = x - p.x;
    var dy:Number = y - p.y;

    var dist:Number = Math.sqrt(dx*dx + dy*dy);

    return dist;
}

Java

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

        float dx = x - x3;
        float dy = y - y3;
        double dist = Math.sqrt(dx*dx + dy*dy);
        return dist;

    }

请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html

按Ctrl +f，输入“segment”，查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。

几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。

WPF版本:

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;
    }
}

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));
}

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;
}