shilvyan  

      /// /****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)
        */

View Code
 1         public double CalOffDistance(GCD gcd, Point3d polylineStartPoint, Point3d polylineEndPoint)
 2         {
 3             double k = (polylineStartPoint.X - polylineEndPoint.X) / (polylineStartPoint.Y - polylineEndPoint.Y);
 4             double c = (polylineStartPoint.Y * polylineEndPoint.X - polylineEndPoint.Y * polylineStartPoint.X) / (polylineStartPoint.Y - polylineEndPoint.Y);
 5             if (polylineStartPoint.Y == polylineEndPoint.Y)
 6             {
 7                 return Math.Abs(gcd.GcdY - polylineEndPoint.Y);
 8             }
 9             return Math.Abs(k * gcd.GcdY - gcd.GcdX + c) / Math.Sqrt(k * k + 1);
10         }

 

posted on 2013-05-10 10:18  shilvyan  阅读(566)  评论(0编辑  收藏  举报