如果它是一条无限大的直线，而不是一条线段，最简单的方法是这样(在ruby中)，其中mx + b是直线，(x1, y1)是已知的点

(y1 - mx1 - b).abs / Math.sqrt(m**2 + 1)

公认的答案行不通 (例如，0,0和(-10,2,10,2)之间的距离应为2)。

下面是工作代码:

   def dist2line2(x,y,line):
     x1,y1,x2,y2=line
     vx = x1 - x
     vy = y1 - y
     ux = x2-x1
     uy = y2-y1
     length = ux * ux + uy * uy
     det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment
     if det < 0:
       return (x1 - x)**2 + (y1 - y)**2
     if det > length:
       return (x2 - x)**2 + (y2 - y)**2
     det = ux * vy - uy * vx
     return det**2 / length
   def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line))

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;

}

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)。

和这个答案一样，只是用的是Visual Basic。使其可作为Microsoft Excel和VBA/宏中的用户定义函数使用。

函数返回点(x,y)到由(x1,y1)和(x2,y2)定义的线段的最近距离。

Function DistanceToSegment(x As Double, y As Double, x1 As Double, y1 As Double, x2 As Double, y2 As Double)

  Dim A As Double
  A = x - x1
  Dim B As Double
  B = y - y1
  Dim C  As Double
  C = x2 - x1
  Dim D As Double
  D = y2 - y1

  Dim dot As Double
  dot = A * C + B * D
  Dim len_sq As Double
  len_sq = C * C + D * D
  Dim param As Double
  param = -1

  If (len_sq <> 0) Then
      param = dot / len_sq
  End If

  Dim xx As Double
  Dim yy As Double

  If (param < 0) Then
    xx = x1
    yy = y1
  ElseIf (param > 1) Then
    xx = x2
    yy = y2
  Else
    xx = x1 + param * C
    yy = y1 + param * D
  End If

  Dim dx As Double
  dx = x - xx
  Dim dy As Double
  dy = y - yy

  DistanceToSegment = Math.Sqr(dx * dx + dy * dy)

End Function

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);
}