GLSL版:

// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
    float aside = dot((p - a),(b - a));
    if(aside< 0.0) return length(p-a);
    float bside = dot((p - b),(a - b));
    if(bside< 0.0) return length(p-b);
    vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
    return length(p - pointOnLine);
}

这是Javascript中最简单的完整代码。

(X, y)是目标点(x1, y)到(x2, y)是线段。

更新:修复了评论中0长度的行问题。

function pDistance(x, y, x1, y1, x2, y2) {

  var A = x - x1;
  var B = y - y1;
  var C = x2 - x1;
  var D = y2 - y1;

  var dot = A * C + B * D;
  var len_sq = C * C + D * D;
  var param = -1;
  if (len_sq != 0) //in case of 0 length line
      param = dot / len_sq;

  var xx, yy;

  if (param < 0) {
    xx = x1;
    yy = y1;
  }
  else if (param > 1) {
    xx = x2;
    yy = y2;
  }
  else {
    xx = x1 + param * C;
    yy = y1 + param * D;
  }

  var dx = x - xx;
  var dy = y - yy;
  return Math.sqrt(dx * dx + dy * dy);
}

更新:Kotlin版本

fun getDistance(x: Double, y: Double, x1: Double, y1: Double, x2: Double, y2: Double): Double {
    val a = x - x1
    val b = y - y1
    val c = x2 - x1
    val d = y2 - y1

    val lenSq = c * c + d * d
    val param = if (lenSq != .0) { //in case of 0 length line
        val dot = a * c + b * d
        dot / lenSq
    } else {
        -1.0
    }

    val (xx, yy) = when {
        param < 0 -> x1 to y1
        param > 1 -> x2 to y2
        else -> x1 + param * c to y1 + param * d
    }

    val dx = x - xx
    val dy = y - yy
    return hypot(dx, dy)
}

以下是Grumdrig解决方案的一个更完整的说明。这个版本还返回最近的点本身。

#include "stdio.h"
#include "math.h"

class Vec2
{
public:
    float _x;
    float _y;

    Vec2()
    {
        _x = 0;
        _y = 0;
    }

    Vec2( const float x, const float y )
    {
        _x = x;
        _y = y;
    }

    Vec2 operator+( const Vec2 &v ) const
    {
        return Vec2( this->_x + v._x, this->_y + v._y );
    }

    Vec2 operator-( const Vec2 &v ) const
    {
        return Vec2( this->_x - v._x, this->_y - v._y );
    }

    Vec2 operator*( const float f ) const
    {
        return Vec2( this->_x * f, this->_y * f );
    }

    float DistanceToSquared( const Vec2 p ) const
    {
        const float dX = p._x - this->_x;
        const float dY = p._y - this->_y;

        return dX * dX + dY * dY;
    }

    float DistanceTo( const Vec2 p ) const
    {
        return sqrt( this->DistanceToSquared( p ) );
    }

    float DotProduct( const Vec2 p ) const
    {
        return this->_x * p._x + this->_y * p._y;
    }
};

// return minimum distance between line segment vw and point p, and the closest point on the line segment, q
float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q )
{
    const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt
    if ( distSq == 0.0 )
    {
        // v == w case
        (*q) = v;

        return v.DistanceTo( p );
    }

    // 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

    const float t = ( p - v ).DotProduct( w - v ) / distSq;
    if ( t < 0.0 )
    {
        // beyond the v end of the segment
        (*q) = v;

        return v.DistanceTo( p );
    }
    else if ( t > 1.0 )
    {
        // beyond the w end of the segment
        (*q) = w;

        return w.DistanceTo( p );
    }

    // projection falls on the segment
    const Vec2 projection = v + ( ( w - v ) * t );

    (*q) = projection;

    return p.DistanceTo( projection );
}

float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY )
{
    Vec2 q;

    float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q );

    (*qX) = q._x;
    (*qY) = q._y;

    return distance;
}

void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY )
{
    float qX;
    float qY;
    float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY );
    printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n",
            segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY );
}

void TestDistanceFromLineSegmentToPoint()
{
    TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 );
}

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)

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)。

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);
}