经过图像处理，得到两直线方程，求两直线交点； 

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)

void GetCrossPoint(float A1, float B1, float A3, float B3,int count,vector<Point2d>& cross_Pt)
{   
    int L1x0 = 1, L1x1 = 500, L3x0 = 1, L3x1 = 500;   //像素坐标X上随便取两点
    float L1y0 = A1*L1x0 + B1;                        //根据已知直线方程求对应的Y值
    float L1y1 = A1*L1x1 + B1;
    float L3y0 = A3*L3x0 + B3;
    float L3y1 = A3*L3x1 + B3;
    float a1 = L1y0 - L1y1;
    float b1 = L1x1 - L1x0;
    float c1 = L1x0*L1y1 - L1x1*L1y0;
    float a3 = L3y0 - L3y1;
    float b3 = L3x1 - L3x0;
    float c3 = L3x0*L3y1 - L3x1*L3y0;
    float D = a1*b3 - a3*b1;
    double x = (b1*c3 - b3*c1) / D;
    double y = (a3*c1 - a1*c3) / D;
    Point2d tmp_pt;
    tmp_pt.x = x;
    tmp_pt.y = y;
    cross_Pt.push_back(tmp_pt);
}

百度知道上找的，先留着慢慢研究：
来源：http://zhidao.baidu.com/question/79460681.html
//.h文件
#ifndef _LINE_H_
#define _LINE_H_

#include <stdio.h>
#include <iostream.h>

class Point
{
public:
 /* 点的坐标 */
 int x1;
 int y1;

public:
 void SetXY(int x, int y);
    void GetXY(int &x, int &y);
 
 /* 2个构造函数 */
 Point();
 Point(int x, int y);
};

class Line : public Point /* line继承point */
{
public:
 /* 另一个点的坐标 */
 int x2;
 int y2;

public:
 void SetPoint(Point* point1, Point* point2);

 /* 3个构造函数 */
 Line();
 Line(int x1, int y1, int x2, int y2);
 Line(Point* point1, Point* point2);

 /* 是否相交 
 ** 相交返回0, 交点为intersect_point
 ** 不相交返回-1, intersect_point为空
 ** 两直线相同返回1, intersect_point为空
 */
 int Intersect(Line* another_line, Point* intersect_point);
};

#endif

//.c文件
#include "1.h"

Point::Point()
{
 x1 = 0;
 y1 = 0;
}

Point::Point(int x, int y)
{
 x1 = x;
 y1 = y;
}

void Point::SetXY(int x, int y)
{
 x1 = x;
 y1 = y;
}


void Point::GetXY(int &x, int &y)
{
 x = x1;
 y = y1;
}

Line::Line()
{
 x1 = 0;
 x2 = 0;
 y1 = 0;
 y2 = 0;
}

Line::Line(int x1, int y1, int x2, int y2)
{
 this->x1 = x1;
 this->x2 = x2;
 this->y1 = y1;
 this->y2 = y2;
}

void Line::SetPoint(Point* point1, Point* point2)
{
 x1 = point1->x1;
 y1 = point1->y1;
 x2 = point2->x1;
    y2 = point2->y1;
}

Line::Line(Point* point1, Point* point2)
{
 x1 = point1->x1;
 y1 = point1->y1;
 x2 = point2->x1;
    y2 = point2->y1;
}

int Line::Intersect(Line* another_line, Point* intersect_point)
{
 /* y = ax + b */
 int a_my, b_my;
 b_my = (x1 * y2 - y1 * x2) / (x1 - x2);
 a_my = (y1 - y2) / (x1 - x2);

 /* check if point */
 Point* point = (Point*)another_line;
 if(a_my * point->x1 + b_my == point->y1)
 {
  intersect_point = point;
  return 0;
 }

 int a_other, b_other;
 a_other = (another_line->x1 * another_line->y2 - another_line->y1 * another_line->x2) / (another_line->x1 - another_line->x2);
 b_other = (another_line->y1 - another_line->y2) / (another_line->x1 - another_line->x2);

 if(a_other == a_my)
 {
  if(b_my == b_other)
  {
   intersect_point = NULL;
   return -1; //not intersect
  }
  else
  {
   return 1; //the same
  }
 }
 else
 {
  intersect_point->x1 = ((b_other - b_my) / (a_my - a_other));
  intersect_point->y1 = (a_my * intersect_point->x1 + b_my);
  return 0; //intersect
 }

}

/* test */
int main()
{
 Point point1(0, 1);
 Point point2(1, 2);

 Line line1(0, 1, 1, 2);
 Line line2(&point1, &point1);

 Line po;
 int a = line1.Intersect((Line*)&line2, &po);
 if(a == 0)
 {
  cout<<"yes"<<endl;
  cout<<"x = "<<po.x1<<", y = "<<po.y1<<endl;
 }
 else if(a == -1)
 {
  cout<<"no"<<endl;
 }
 else
 {
  cout<<"same"<<endl;
 }

 return 0;
}
 

假设两直线的式子分别为：



求解过程: 





















 综上所述，交点的解为：



 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：



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;
}

 

已知点A（x1，y1）、B（x2，y2）为直线L1上两个点，点C（x3，y3）、D（x4，y4）为直线L2上两个点，求解两个直线的方程以及交点坐标，直接贴代码

#include <iostream>

using namespace std;


//定义Point2f结构体  
struct Point2f
{
	float x;
	float y;
};

// 定义直线参数结构体  
struct LinePara
{
	float k;
	float b;

};

// 获取直线参数  
void getLinePara(float& x1, float& y1, float& x2, float& y2, LinePara & LP)
{
	double m = 0;

	// 计算分子  
	m = x2 - x1;

	if (0 == m)
	{
		LP.k = 10000.0;
		LP.b = y1 - LP.k * x1;
	}
	else
	{
		LP.k = (y2 - y1) / (x2 - x1);
		LP.b = y1 - LP.k * x1;
	}


}

// 获取交点  
bool getCross(float & x1, float &y1, float & x2, float & y2, float & x3, float &y3, float & x4, float & y4,  Point2f & pt){

	LinePara para1, para2;
	getLinePara(x1, y1, x2, y2, para1);
	getLinePara(x3, y3, x4, y4, para2);

	// 判断是否平行  
	if (abs(para1.k - para2.k) > 0.5)
	{
		pt.x = (para2.b - para1.b) / (para1.k - para2.k);
		pt.y = para1.k * pt.x + para1.b;

		return true;

	}
	else
	{
		return false;
	}

}

void main()
{
	Point2f pt1, pt2, pt3, pt4, pt;
	pt1.x = 1.0;
	pt1.y = 1.0;

	pt2.x = 2.0;
	pt2.y = 2.0;

	pt3.x = 0.0;
	pt3.y = 2.0;

	pt4.x = 2.0;
	pt4.y = 0.0;

	getCross(pt1.x, pt1.y, pt2.x, pt2.y, pt3.x, pt3.y, pt4.x, pt4.y, pt);

	cout << pt.x << " , " << pt.y << endl;
}


直接来个完整版的，ctrl+F5，运行效果如下，打完收工！
在这里感谢大佬给的建议，将int m改成double m，谢谢~