计算两条直线的交点(C#)

/// <summary>
/// 计算两条直线的交点
/// </summary>
/// <param name="lineFirstStar">L1的点1坐标</param>
/// <param name="lineFirstEnd">L1的点2坐标</param>
/// <param name="lineSecondStar">L2的点1坐标</param>
/// <param name="lineSecondEnd">L2的点2坐标</param>
/// <returns></returns>
public static PointF GetIntersection(PointF lineFirstStar, PointF lineFirstEnd, PointF lineSecondStar, PointF lineSecondEnd)
{
/*
* L1，L2都存在斜率的情况：
* 直线方程L1: ( y - y1 ) / ( y2 - y1 ) = ( x - x1 ) / ( x2 - x1 )
* => y = [ ( y2 - y1 ) / ( x2 - x1 ) ]( x - x1 ) + y1
* 令 a = ( y2 - y1 ) / ( x2 - x1 )
* 有 y = a * x - a * x1 + y1 .........1
* 直线方程L2: ( y - y3 ) / ( y4 - y3 ) = ( x - x3 ) / ( x4 - x3 )
* 令 b = ( y4 - y3 ) / ( x4 - x3 )
* 有 y = b * x - b * x3 + y3 ..........2
*
* 如果 a = b，则两直线平等，否则，联解方程 1,2，得:
* x = ( a * x1 - b * x3 - y1 + y3 ) / ( a - b )
* y = a * x - a * x1 + y1
*
* L1存在斜率, L2平行Y轴的情况：
* x = x3
* y = a * x3 - a * x1 + y1
*
* L1 平行Y轴，L2存在斜率的情况：
* x = x1
* y = b * x - b * x3 + y3
*
* L1与L2都平行Y轴的情况：
* 如果 x1 = x3，那么L1与L2重合，否则平等
*
*/
float a = 0, b = 0;
int state = 0;
if (lineFirstStar.X != lineFirstEnd.X)
{
a = (lineFirstEnd.Y - lineFirstStar.Y) / (lineFirstEnd.X - lineFirstStar.X);
state |= 1;
}
if (lineSecondStar.X != lineSecondEnd.X)
{
b = (lineSecondEnd.Y - lineSecondStar.Y) / (lineSecondEnd.X - lineSecondStar.X);
state |= 2;
}
switch (state)
{
case 0: //L1与L2都平行Y轴
{
if (lineFirstStar.X == lineSecondStar.X)
{
//throw new Exception("两条直线互相重合，且平行于Y轴，无法计算交点。");
return new PointF(0, 0);
}
else
{
//throw new Exception("两条直线互相平行，且平行于Y轴，无法计算交点。");
return new PointF(0, 0);
}
}
case 1: //L1存在斜率, L2平行Y轴
{
float x = lineSecondStar.X;
float y = (lineFirstStar.X - x) * (-a) + lineFirstStar.Y;
return new PointF(x, y);
}
case 2: //L1 平行Y轴，L2存在斜率
{
float x = lineFirstStar.X;
//网上有相似代码的，这一处是错误的。你可以对比case 1 的逻辑进行分析
//源code:lineSecondStar * x + lineSecondStar * lineSecondStar.X + p3.Y;
float y = (lineSecondStar.X - x) * (-b) + lineSecondStar.Y;
return new PointF(x, y);
}
case 3: //L1，L2都存在斜率
{
if (a == b)
{
// throw new Exception("两条直线平行或重合，无法计算交点。");
return new PointF(0, 0);
}
float x = (a * lineFirstStar.X - b * lineSecondStar.X - lineFirstStar.Y + lineSecondStar.Y) / (a - b);
float y = a * x - a * lineFirstStar.X + lineFirstStar.Y;
return new PointF(x, y);
}
}
// throw new Exception("不可能发生的情况");
return new PointF(0, 0);
}

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计算两条直线的交点(C#)

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