嘿，我昨天才写的。它在Actionscript 3.0中，基本上是Javascript，尽管你可能没有相同的Point类。

//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.  
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
    var nearestPt:Point; //closest point on seqment to pt

    var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
    var bLenSq:Number = dot( b, b ); //Segment length squared

    if( keyDot <= 0 )  //pt is "behind" st, use st
    {
        nearestPt = st  
    }
    else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
    {
        nearestPt = st.add(b);
    }
    else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
    {
        var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
        var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
        nearestPt = st.add(partOfB);
    }

    var dist:Number = (pt.subtract(nearestPt)).length;

    return dist;
}

此外，这里有一个关于这个问题的相当完整和可读的讨论:notejot.com

一个2D和3D的解决方案

考虑基底的变化，使得线段变成(0,0,0)-(d, 0,0)和点(u, v, 0)。在这个平面上，最短的距离由

    u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u     -> d(B, C)

(到其中一个端点或到支撑线的距离，取决于到该线的投影。等距轨迹由两个半圆和两条线段组成。)

式中，d为AB线段的长度，u、v分别为AB/d (AB方向的单位矢量)与AC的标量积和外积的模量。

AB.AC ≤ 0             -> |AC|
    0 ≤ AB.AC ≤ AB²   -> |ABxAC|/|AB|
          AB² ≤ AB.AC -> |BC|

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)

在数学

它使用线段的参数描述，并将点投影到线段定义的直线中。当参数在线段内从0到1时，如果投影在这个范围之外，我们计算到相应端点的距离，而不是法线到线段的直线。

Clear["Global`*"];
 distance[{start_, end_}, pt_] := 
   Module[{param},
   param = ((pt - start).(end - start))/Norm[end - start]^2; (*parameter. the "."
                                                       here means vector product*)

   Which[
    param < 0, EuclideanDistance[start, pt],                 (*If outside bounds*)
    param > 1, EuclideanDistance[end, pt],
    True, EuclideanDistance[pt, start + param (end - start)] (*Normal distance*)
    ]
   ];  

策划的结果:

Plot3D[distance[{{0, 0}, {1, 0}}, {xp, yp}], {xp, -1, 2}, {yp, -1, 2}]

画出比截断距离更近的点:

等高线图:

这是我最后写的代码。这段代码假设一个点以{x:5, y:7}的形式定义。注意，这不是绝对最有效的方法，但它是我能想到的最简单、最容易理解的代码。

// a, b, and c in the code below are all points

function distance(a, b)
{
    var dx = a.x - b.x;
    var dy = a.y - b.y;
    return Math.sqrt(dx*dx + dy*dy);
}

function Segment(a, b)
{
    var ab = {
        x: b.x - a.x,
        y: b.y - a.y
    };
    var length = distance(a, b);

    function cross(c) {
        return ab.x * (c.y-a.y) - ab.y * (c.x-a.x);
    };

    this.distanceFrom = function(c) {
        return Math.min(distance(a,c),
                        distance(b,c),
                        Math.abs(cross(c) / length));
    };
}

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 );
}