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/****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;

        }

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