求两直线的交点（C++）

假设两直线的式子分别为：

$\left\{\begin{matrix} L1:A_{1}x + B_{1}y + C_{1} = 0 \\ L2:A_{2}x + B_{2}y + C_{2} = 0 \end{matrix}\right.$

求解过程:

$(1) L_{1} * A_{2} - L_{2} * A_{1}$

$L_{1} * A_{2} \Rightarrow A_{1} A_{2} x + A_{2} B_{1} y + A_{2} C_{1} = 0;$

$L_{2} * A_{1} \Rightarrow A_{1} A_{2} x + A_{1} B_{2} y + A_{1} C_{2} = 0;$

$L_{1} * A_{2} - L_{2} * A_{1} \Rightarrow A_{2} B_{1} y - A_{1} B_{2} y + A_{2} C_{1} - A_{1} C_{2} = 0$

$\Rightarrow y = (A_{1} C_{2} - A_{2} C_{1} ) / (A_{2} B_{1} - A_{1} B_{2}).$

$(2)L_{1} * B_{2} - L_{2} * B_{1}$

$L_{1} * B_{2} \Rightarrow A_{1} B_{2} x + B_{1} B_{2} y + B_{2} C_{1} = 0;$

$L_{2} * B_{1} \Rightarrow A_{2} B_{1} x + B_{1} B_{2} y + B_{1} C_{2} = 0;$

$L_{1} * B_{2} - L_{2} * B_{1} \Rightarrow A_{1} B_{2} x - A_{2} B_{1} x + B_{2} C_{1} - B_{1} C_{2} = 0$

$\Rightarrow x = (B_{2} C_{1} - B_{1} C_{2}) / (A_{2} B_{1} - A_{1} B_{2}).$

$(3)$ 综上所述，交点的解为：

$\left\{\begin{matrix} x = (B_{2} C_{1} - B_{1} C_{2}) / (A_{2}B_{1} - A_{1} B_{2}) \\ y = (A_{1} C_{2} - A_{2} C_{1}) / (A_{2} B_{1} - A_{1} B_{2}) \end{matrix}\right.$

$(A2 B1 - A1 B2) \neq 0$ 则有解，否则两直线平行。

补充：如果化成Y=kX+b的形式的话，得解为：

$\left\{\begin{matrix}x=(b_{2}-b_{1})/(k_{1}-k_{2}) \\ y = (k_{1}b_{2}-k_{2}b_{1})/(k_{1}-k_{2}) \end{matrix}\right.$

例题：Audio

给定三个不共线的点，要求输出一个点使得这个点到三个点的距离相同，输出保留三位小数。

#include<bits/stdc++.h>
using namespace std;int main()
{double x1,y1,x2,y2,x3,y3;scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);double k1=(y1-y2)/(x1-x2),k2=(y1-y3)/(x1-x3);k1=-1/k1,k2=-1/k2;double x=(x1+x2)/2,y=(y1+y2)/2;double b1=y-k1*x;x=(x1+x3)/2,y=(y1+y3)/2;double b2=y-k2*x;double X=(b2-b1)/(k1-k2);double Y=(k1*b2-k2*b1)/(k1-k2);printf("%.3f %.3f\n",X,Y);return 0;
}