求连直线的焦点

拼搏现实的明天。 2022-05-30 01:29 107阅读 0赞

** 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.csdn.net/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))));  
                }  
        }

![SouthEast][]

本文转载自http://blog.csdn.net/gongwenkai90/article/details/38519623

[SouthEast]: /images/20220530/712944089a524b098dacf80dd635998c.png