/****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;
}