这个答案是基于公认答案的JavaScript解决方案。 它主要只是格式更好，函数名更长，当然函数语法更短，因为它是在ES6 + CoffeeScript中。

JavaScript版本(ES6)

distanceSquared = (v, w)=> Math.pow(v.x - w.x, 2) + Math.pow(v.y - w.y, 2);
distance = (v, w)=> Math.sqrt(distanceSquared(v, w));

distanceToLineSegmentSquared = (p, v, w)=> {
    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)=> {
    return Math.sqrt(distanceToLineSegmentSquared(p, v));
}

CoffeeScript版本

distanceSquared = (v, w)-> (v.x - w.x) ** 2 + (v.y - w.y) ** 2
distance = (v, w)-> Math.sqrt(distanceSquared(v, w))

distanceToLineSegmentSquared = (p, v, w)->
    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)->
    Math.sqrt(distanceToLineSegmentSquared(p, v, w))

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;
}

C#

改编自@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);
}

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。

只是遇到了这个，我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出，直线或线段由包含两个点的表给出(见下面的例子):

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 >= 0 and t <= 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

使用示例:

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

JavaScript中一个基于这个公式的更简洁的解决方案:

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;

}