计算有一条垂直线的交点

Evaluating the intersection point with a vertical line involved

本文关键字：垂直线 一条 计算      更新时间：2023-10-16

函数返回由点p0, p1和p2, p3定义的两条直线之间的交点:

#include "std_lib_facilities.h"
#include <iostream>
struct Point{
    Point(int xx, int yy): x(xx), y(yy) { }
int x;
int y;
};
// find the intersection of two lines
Point intersectPoint(Point& p0, Point& p1, Point& p2, Point& p3){
    // line formed by p0p1
    double dx1 = p1.x - p0.x;
    double dy1 = p1.y - p0.y;
    double m1 = dy1 / dx1;
    double c1 = p1.y - m1 * p1.x;
    cout <<"m1: "<< m1 <<'n'; 
    // line formed by p2p3
    double dx2 = p3.x - p2.x;
    double dy2 = p3.y - p2.y;
    double m2 = dy2 / dx2;
    double c2 = p3.y - m1 * p3.x;
    cout <<"m2: "<< m2 <<'n';
    // find intersection point
    double epsilon = 1e-6;
    // (-1,-1) represents no intersection 
    if (abs(m1 - m2) < epsilon) return Point(-1,-1);
    else{
       double interX = (c2 - c1) / (m1 - m2);
       double interY = m1 * interX + c1;
       cout <<"(interX, interY): "<< interX <<", "<< interY <<'n';
       return Point(interX, interY);
   }
}
//------------------------------------------------------------------------------------------------
int main(){ 
    Point verticalStart(100, 100);
    Point verticalEnd(100,300);
    Point horizontalStart(50,200);
    Point horizontalEnd(150,200);
    Point intersection = intersectPoint(horizontalStart ,horizontalEnd, verticalStart, verticalEnd);
    cout <<"returned value: n"; 
    cout << "( "<< intersection.x <<", "<< intersection.y <<" )n"; 
    getchar();
    return 0;
}

在垂直和水平相交的情况下，我得到一个输出:

m1: 0
m2: 1.#inf
(interX, interY): -0, 200
returned value:
( 0, 200 )

问题:

如何处理这种情况，使函数返回有效的交点?

注意:代码是在MVS2010上编译的

我建议不要使用斜截表示，即:y = a*x + b，而是矢量参数表示:

Pa = (P1-P0)*t + P0
Pb = (P3-P2)*u + P2

这样，行x = k确实有一个表示。

对变量t和u(两个变量，两个方程，对于两个坐标x和y)求解方程组，其中Pa = Pb(交点)，并在其中一个方程中反代得到交点。

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