求两条直线的交点，运用面向对象的思想编程实现C++源码

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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

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  6class 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))));}}

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