点到直线的距离

 c#版本

         /****点到直线的距离***

         * 过点（x1,y1）和点（x2,y2）的直线方程为：KX -Y + (x2y1 - x1y2)/(x2-x1) = 0
         * 设直线斜率为K = (y2-y1)/(x2-x1),C=(x2y1 - x1y2)/(x2-x1)
         * 点P(x0,y0)到直线AX + BY +C =0DE 距离为：d=|Ax0 + By0 + C|/sqrt(A*A + B*B)
         * 点（x3,y3）到经过点（x1,y1）和点（x2,y2）的直线的最短距离为：
         * distance = |K*x3 - y3 + C|/sqrt(K*K + 1)
         */
        public static double GetMinDistance(IPoint pt1, IPoint pt2, IPoint pt3)
        {
            double dis = 0;
            if (pt1.X == pt2.X)
            {
                dis = Math.Abs(pt3.X - pt1.X);
                return dis;
            }
            double lineK = (pt2.Y - pt1.Y) / (pt2.X - pt1.X);
            double lineC = (pt2.X * pt1.Y - pt1.X * pt2.Y) / (pt2.X - pt1.X);
            dis = Math.Abs(lineK * pt3.X - pt3.Y + lineC) / (Math.Sqrt(lineK * lineK + 1));
            return dis;

        }

 VB版本

Public Function GetVerticalPoint(ByVal x1 As Double, ByVal y1 As Double, ByVal x2 As Double, ByVal y2 As Double, ByVal x3 As Double, ByVal y3 As Double, ByRef x As Double, ByRef y As Double) As Boolean
'(x,y)返回垂足;(x1,y1)为测试点;(x2,y2)(x3,y3)为直线点
On Error GoTo PROC_ERROR
    Dim result As New MapXLib.Point
    If x2 = x3 Then '垂直线
        x = x2
        y = y1
        GetVerticalPoint = True
        Exit Function
    Else
        Dim kk As Double
        kk = (y3 - y2) / (x3 - x2)
        x = (y1 - y2 + kk * x2 + (1 / kk) * x1) / ((1 / kk) + kk)
        y = kk * x - x2 * kk + y2
        GetVerticalPoint = True
        Exit Function
    End If
    GetVerticalPoint = False
    Exit Function
PROC_ERROR:
    MsgBox Err.Description
End Function