||
c# 实现点到直线的距离
/****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(PointF pt1, PointF pt2, PointF 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;
} =================================== 使用海伦公式==================================== double dPtToLine = new double();
double a, b, c;
a = DPtToPtWin(pt0, pt1);// 线段的长度
b = DPtToPtWin(pt0, pt);// Pt1到点的距离 c = DPtToPtWin(pt1, pt);// Pt2到点的距离 #region 两点屏幕距离
public double DPtToPtWin(PointF pt0, PointF pt1)
{
return Math.Sqrt(Math.Pow((pt0.X - pt1.X), 2) + Math.Pow((pt0.Y - pt1.Y), 2));
} #endregion double p = (a + b + c) / 2;// 半周长
double s = Math.Sqrt(p * (p - a) * (p - b) * (p - c));// 海伦公式求面积 dPtToLine = 2 * s / a;// 返回点到线的距离(利用三角形面积公式求高) |
Archiver|手机版|科学网 ( 京ICP备07017567号-12 )
GMT+8, 2024-5-14 12:25
Powered by ScienceNet.cn
Copyright © 2007- 中国科学报社