点到直线的距离

 /****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(double x,double y,double x1,double y1,double x2,double y2)
        {
            double dis = 0;
            if (Math.Abs(x1- x2)<0.00001)
            {
                dis = Math.Abs(x - x1);
                return dis;
            }
            double lineK = (y2 - y1) / (x2 - x1);//斜率
            double lineC = (x2 * y1 - x1 * y2) / (x2 - x1);
            dis = Math.Abs(lineK * x - y + lineC) / (Math.Sqrt(lineK * lineK + 1));
            return dis;

        }