这里没有看到Java实现，所以我将Javascript函数从接受的答案转换为Java代码:

static double sqr(double x) {
    return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
    return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
    double l2 = dist2(v, w);
    if (l2 == 0) return dist2(p, v);
    double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    if (t < 0) return dist2(p, v);
    if (t > 1) return dist2(p, w);
    return dist2(p, new DoublePoint(
            v.x + t * (w.x - v.x),
            v.y + t * (w.y - v.y)
    ));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
    return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
    public double x;
    public double y;

    public DoublePoint(double x, double y) {
        this.x = x;
        this.y = y;
    }
}

在f#中，点c到a和b之间的线段的距离为:

let pointToLineSegmentDistance (a: Vector, b: Vector) (c: Vector) =
  let d = b - a
  let s = d.Length
  let lambda = (c - a) * d / s
  let p = (lambda |> max 0.0 |> min s) * d / s
  (a + p - c).Length

向量d沿着线段从a指向b。d/s与c-a的点积给出了无限直线与点c之间最接近点的参数。使用min和max函数将该参数钳制到范围0..s，使该点位于a和b之间。最后，a+p-c的长度是c到线段上最近点的距离。

使用示例:

pointToLineSegmentDistance (Vector(0.0, 0.0), Vector(1.0, 0.0)) (Vector(-1.0, 1.0))

对于懒人来说，以下是我在Objective-C语言中移植@Grumdrig的解决方案:

CGFloat sqr(CGFloat x) { return x*x; }
CGFloat dist2(CGPoint v, CGPoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); }
CGFloat distanceToSegmentSquared(CGPoint p, CGPoint v, CGPoint w)
{
    CGFloat l2 = dist2(v, w);
    if (l2 == 0.0f) return dist2(p, v);

    CGFloat t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    if (t < 0.0f) return dist2(p, v);
    if (t > 1.0f) return dist2(p, w);
    return dist2(p, CGPointMake(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y)));
}
CGFloat distanceToSegment(CGPoint point, CGPoint segmentPointV, CGPoint segmentPointW)
{
    return sqrtf(distanceToSegmentSquared(point, segmentPointV, segmentPointW));
}

本想在GLSL中这样做，但如果可能的话，最好避免所有这些条件。使用clamp()可以避免两种端点情况:

// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
    vec3 AP = P - A, AB = B - A;
    float l = dot(AB, AB);
    if (l <= 0.0000001) return A;    // A and B are practically the same
    return AP - AB*clamp(dot(AP, AB)/l, 0.0, 1.0);  // do the projection
}

如果您可以确定A和B彼此不会非常接近，则可以简化为删除If()。事实上，即使A和B是相同的，我的GPU仍然给出了这个无条件版本的正确结果(但这是使用pre-OpenGL 4.1，其中GLSL除零是未定义的):

// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
    vec3 AP = P - A, AB = B - A;
    return AP - AB*clamp(dot(AP, AB)/dot(AB, AB), 0.0, 1.0);
}

计算距离是很简单的——GLSL提供了一个distance()函数，你可以在这个最近的点和P。

灵感来自Iñigo Quilez的胶囊距离函数代码

这里它使用Swift

    /* Distance from a point (p1) to line l1 l2 */
func distanceFromPoint(p: CGPoint, toLineSegment l1: CGPoint, and l2: CGPoint) -> CGFloat {
    let A = p.x - l1.x
    let B = p.y - l1.y
    let C = l2.x - l1.x
    let D = l2.y - l1.y

    let dot = A * C + B * D
    let len_sq = C * C + D * D
    let param = dot / len_sq

    var xx, yy: CGFloat

    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
    }

    let dx = p.x - xx
    let dy = p.y - yy

    return sqrt(dx * dx + dy * dy)
}

下面是HSQLDB的SQL实现:

CREATE FUNCTION dist_to_segment(px double, py double, vx double, vy double, wx double, wy double)
  RETURNS double
BEGIN atomic
   declare l2 double;
   declare t double;
   declare nx double;
   declare ny double;
   set l2 =(vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
   IF l2 = 0 THEN
     RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
   ELSE
     set t = ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
     set t = GREATEST(0, LEAST(1, t));
     set nx=vx + t * (wx - vx);
     set ny=vy + t * (wy - vy);
     RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
   END IF;
END;

Postgres的实现:

CREATE FUNCTION dist_to_segment(px numeric, py numeric, vx numeric, vy numeric, wx numeric, wy numeric)
  RETURNS numeric
AS $$
   declare l2 numeric;
   declare t numeric;
   declare nx numeric;
   declare ny numeric;
BEGIN 
   l2 := (vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
   IF l2 = 0 THEN
     RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
   ELSE
     t := ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
     t := GREATEST(0, LEAST(1, t));
     nx := vx + t * (wx - vx);
     ny := vy + t * (wy - vy);
     RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
   END IF;
END;
$$ LANGUAGE plpgsql;