求连直线的焦点

 一般方程法：

直线的一般方程为F(x) = ax + by + c = 0。既然我们已经知道直线的两个点，假设为(x0,y0), (x1, y1)，那么可以得到a = y0 – y1, b = x1 – x0, c = x0y1 – x1y0。

因此我们可以将两条直线分别表示为

F0(x) = a0*x + b0*y + c0 = 0, F1(x) = a1*x + b1*y + c1 = 0

那么两条直线的交点应该满足

a0*x + b0*y +c0 = a1*x + b1*y + c1

由此可推出

x = (b0*c1 – b1*c0)/D

y = (a1*c0 – a0*c1)/D

D = a0*b1 – a1*b0， (D为0时，表示两直线平行)

二者实际上就是连立方程组F0(x) = a0*x + b0*y + c0 = 0, F1(x) = a1*x + b1*y + c1 = 0的叉积应用

i     j     k

a0 b0 c0

a1 b1 c1

a = y0 – y1, b = x1 – x0, c = x0y1 – x1y0推盗计算过程：

y=kx+m,
y1=kx1+m
y2=kx2+m,
k=(y2-y1)/(x2-x1)
m=y1-(y2-y1)*x1/(x2-x1)
y=(y2-y1)/(x2-x1)x+(y1*(x2-x1)-x1(y2-y1))/(x2-x1)
(y1-y2)x+(x2-x1)y+(x1y2-x2y1)=0,
a=y1-y2,
b=x2-x1,
c=x1y2-x2y1,

此方法摘自 http://blog.youkuaiyun.com/abcjennifer/article/details/7584628，本人亲自推演过。

    #include<iostream>  
    #include<iomanip>  
    using namespace std;  
      
    #define N  6  
    class Point{  
    public:  
        double m_pointX;  
        double m_pointY;  
    public:  
        Point(){}  
        Point(double x, double y){m_pointX = x; m_pointY = y;}  
    };  
      
    class Line:public Point{  
    public:  
        double a;  
        double b;  
        double c;  
    public:  
        Line GetLine(Point ptSource, Point ptDestination);  
        Point GetCrossPoint(Line l1, Line l2);  
        void CrossPointShow(Point ptCross);  
    };  
      
    Line Line::GetLine(Point ptSource, Point ptDestination)  
    {  
        Line lTemp;  
        lTemp.a = ptSource.m_pointY - ptDestination.m_pointY;    
        lTemp.b = ptDestination.m_pointX - ptSource.m_pointX;    
        lTemp.c = ptSource.m_pointX*ptDestination.m_pointY - ptDestination.m_pointX*ptSource.m_pointY;  
        return lTemp;  
    }  
      
    Point Line::GetCrossPoint(Line l1, Line l2)  
    {  
        Point pTemp;  
        double D;  
        D = l1.a*l2.b - l2.a*l1.b;    
        Point p;    
        pTemp.m_pointX = (l1.b*l2.c - l2.b*l1.c)/D;    
        pTemp.m_pointY = (l1.c*l2.a - l2.c*l1.a)/D;    
        return pTemp;  
    }  
      
    void Line::CrossPointShow(Point ptCross)  
    {  
        cout<<"两条直线交点的横坐标："<<setprecision(N)<<ptCross.m_pointX<<endl;  
        cout<<"两条直线交点的纵坐标："<<setprecision(N)<<ptCross.m_pointY<<endl;  
    }  
      
    void main()  
    {  
            Line l;  
            double x0, x1, x2, x3, y0, y1, y2, y3;  
            char c0, c1, d0;  
            while (1)  
            {  
                cout<<"请输入一条直线的两点坐标："<<endl;  
                cin>>c0>>x0>>d0>>y0>>c1>>c0>>x1>>d0>>y1>>c1;  
                cout<<"请输入另一条直线的两点坐标："<<endl;  
                cin>>c0>>x2>>d0>>y2>>c1>>c0>>x3>>d0>>y3>>c1;  
                l.CrossPointShow(l.GetCrossPoint(l.GetLine(Point(x0, y0), Point(x1, y1)), l.GetLine(Point(x2, y2), Point(x3, y3))));  
            }  
    }  

本文转载自http://blog.youkuaiyun.com/gongwenkai90/article/details/38519623