arengine 点到直线的距离

 

点到直线的距离

/****点到直线的距离***
         * 2007-11-20
         * Robbin
         * 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;
        }

来自:http://www.cnblogs.com/lauer0246/archive/2009/06/24/1510363.html