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)

使用arctangents的一行解决方案:

思路是将A移动到(0,0)，并顺时针旋转三角形，使C位于X轴上， 当这种情况发生时，By就是距离。

a角= Atan(Cy - Ay, Cx - Ax); b角= Atan(By - Ay, Bx - Ax); AB长度=平方根((Bx - Ax)²+ (By - Ay)²) By = Sin (bAngle - aAngle) * ABLength

C#

public double Distance(Point a, Point b, Point c)
{
    // normalize points
    Point cn = new Point(c.X - a.X, c.Y - a.Y);
    Point bn = new Point(b.X - a.X, b.Y - a.Y);

    double angle = Math.Atan2(bn.Y, bn.X) - Math.Atan2(cn.Y, cn.X);
    double abLength = Math.Sqrt(bn.X*bn.X + bn.Y*bn.Y);

    return Math.Sin(angle)*abLength;
}

一行c#(要转换为SQL)

double distance = Math.Sin(Math.Atan2(b.Y - a.Y, b.X - a.X) - Math.Atan2(c.Y - a.Y, c.X - a.X)) * Math.Sqrt((b.X - a.X) * (b.X - a.X) + (b.Y - a.Y) * (b.Y - a.Y))

忍不住用python来编码:)

from math import sqrt, fabs
def pdis(a, b, c):
    t = b[0]-a[0], b[1]-a[1]           # Vector ab
    dd = sqrt(t[0]**2+t[1]**2)         # Length of ab
    t = t[0]/dd, t[1]/dd               # unit vector of ab
    n = -t[1], t[0]                    # normal unit vector to ab
    ac = c[0]-a[0], c[1]-a[1]          # vector ac
    return fabs(ac[0]*n[0]+ac[1]*n[1]) # Projection of ac to n (the minimum distance)

print pdis((1,1), (2,2), (2,0))        # Example (answer is 1.414)

fortran也是一样:)

real function pdis(a, b, c)
    real, dimension(0:1), intent(in) :: a, b, c
    real, dimension(0:1) :: t, n, ac
    real :: dd
    t = b - a                          ! Vector ab
    dd = sqrt(t(0)**2+t(1)**2)         ! Length of ab
    t = t/dd                           ! unit vector of ab
    n = (/-t(1), t(0)/)                ! normal unit vector to ab
    ac = c - a                         ! vector ac
    pdis = abs(ac(0)*n(0)+ac(1)*n(1))  ! Projection of ac to n (the minimum distance)
end function pdis


program test
    print *, pdis((/1.0,1.0/), (/2.0,2.0/), (/2.0,0.0/))   ! Example (answer is 1.414)
end program test

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));
    };
}

我制作了一个交互式Desmos图来演示如何实现这一点:

https://www.desmos.com/calculator/kswrm8ddum

红点是A点，绿点是B点，C点是蓝色点。 您可以拖动图形中的点来查看值的变化。 左边的值“s”是线段的参数(即s = 0表示点A, s = 1表示点B)。 值“d”是第三点到经过A和B的直线的距离。

编辑:

有趣的小见解:坐标(s, d)是坐标系中第三点C的坐标，AB是单位x轴，单位y轴垂直于AB。

快速实现http://paulbourke.net/geometry/pointlineplane/source.c

    static func magnitude(p1: CGPoint, p2: CGPoint) -> CGFloat {
        let vector = CGPoint(x: p2.x - p1.x, y: p2.y - p1.y)
        return sqrt(pow(vector.x, 2) + pow(vector.y, 2))
    }

    /// http://paulbourke.net/geometry/pointlineplane/
    /// http://paulbourke.net/geometry/pointlineplane/source.c
    static func pointDistanceToLine(point: CGPoint, lineStart: CGPoint, lineEnd: CGPoint) -> CGFloat? {

        let lineMag = magnitude(p1: lineEnd, p2: lineStart)
        let u = (((point.x - lineStart.x) * (lineEnd.x - lineStart.x)) +
                ((point.y - lineStart.y) * (lineEnd.y - lineStart.y))) /
                (lineMag * lineMag)

        if u < 0 || u > 1 {
            // closest point does not fall within the line segment
            return nil
        }

        let intersectionX = lineStart.x + u * (lineEnd.x - lineStart.x)
        let intersectionY = lineStart.y + u * (lineEnd.y - lineStart.y)

        return magnitude(p1: point, p2: CGPoint(x: intersectionX, y: intersectionY))
    }