求直线和直线交点

/// <summary>
/// 求直线和直线交点
/// </summary>
public double[] GetIntersection(double x1, double y1, double x2, double y2, double x3, double y3, double x4, double y4)
{

//取得交点
var a = (y2 - y1) / (x2 - x1); //需考虑分母不能为0 即x2=x1 l1垂直于x轴
var b = (y4 - y3) / (x4 - x3); //需考虑分母不能为0 即x4=x3 l2垂直于x轴

if (a == b)
{//斜率相同,说明平行 无交点
//throw new Exception("两直线平行,无交点");
return null;
}

double x, y = 0;

if (x2 == x1)
{//L1垂直于x轴 则x=x1=x2 a=infinity 想办法消除a
x = x1;
////(y-y3)/(x-x3)=b 且x=x1 变换得y=bx1-bx3+y3
y = b * x1 - b * x3 + y3;
return new double[] { x, y };
}
if (x4 == x3)
{//L2垂直于x轴 则x=x3=x4 b=infinity
x = x3;
y = a * x - a * x1 + y1;
return new double[] { x, y };
}

x = (a * x1 - y1 + y3 - b * x3) / (a - b);
y = a * x - a * x1 + y1;

 

//前面算法自动延伸了

//判断交点是否在直线上

double dSquareSum = 0;
dSquareSum = Math.Pow(x2 - x1, 2) + Math.Pow(y2 - y1, 2);
double len3 = Math.Sqrt(dSquareSum);

dSquareSum = 0;
dSquareSum = Math.Pow(x - x1, 2) + Math.Pow(y - y1, 2);
double len1 = Math.Sqrt(dSquareSum);

dSquareSum = 0;
dSquareSum = Math.Pow(x - x2, 2) + Math.Pow(y - y2, 2);
double len2 = Math.Sqrt(dSquareSum);

dSquareSum = 0;
dSquareSum = Math.Pow(x - x3, 2) + Math.Pow(y - y3, 2);
double len4 = Math.Sqrt(dSquareSum);

dSquareSum = 0;
dSquareSum = Math.Pow(x - x4, 2) + Math.Pow(y - y4, 2);
double len5 = Math.Sqrt(dSquareSum);

dSquareSum = 0;
dSquareSum = Math.Pow(x4 - x3, 2) + Math.Pow(y4 - y3, 2);
double len6 = Math.Sqrt(dSquareSum);

if ((len1 + len2 - len3) < 0.1 && (len4 + len5 - len6) < 0.1)
{
return new double[] { Math.Round(x,2), Math.Round(y,2) };
}
else
{
return null;
}

}