Intersecting Lines

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect. 
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000. 
Input
The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).
Output
There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".
Sample Input
5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5
Sample Output
INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT



#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
//#include<bits/stdc++.h>
#define eps 1e-8
using namespace std;
int main()
{
    int n;
    double x1,y1,x2,y2,x3,y3,x4,y4;
    scanf("%d",&n);
    printf("INTERSECTING LINES OUTPUT\n");
    for(int i=1;i<=n;i++)
    {
      scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
       if((x4-x3)*(y2-y1)==(y4-y3)*(x2-x1))//斜率相同
       {
           if((x2-x1)*(y3-y1)==(y2-y1)*(x3-x1))
               printf("LINE\n");
          else
               printf("NONE\n");
       }
       else
       {
            double a1=y1-y2,b1=x2-x1,c1=x1*y2-x2*y1;
            double a2=y3-y4,b2=x4-x3,c2=x3*y4-x4*y3;
            double x=(c2*b1-c1*b2)/(b2*a1-b1*a2);
            double y=(a2*c1-a1*c2)/(b2*a1-b1*a2);
            printf("POINT %.2f %.2f\n",x,y);

       }
    }
    printf("END OF OUTPUT\n");
}




Here's one possible solution in Java: ```java import java.util.Scanner; public class IntersectingPoint { public static void main(String[] args) { // Prompt the user to enter the four endpoints Scanner input = new Scanner(System.in); System.out.print("Enter x1, y1, x2, y2, x3, y3, x4, y4: "); double x1 = input.nextDouble(); double y1 = input.nextDouble(); double x2 = input.nextDouble(); double y2 = input.nextDouble(); double x3 = input.nextDouble(); double y3 = input.nextDouble(); double x4 = input.nextDouble(); double y4 = input.nextDouble(); // Create two LinearEquation objects for the two line segments LinearEquation eq1 = new LinearEquation(y1 - y2, x2 - x1, y1*x2 - y2*x1); LinearEquation eq2 = new LinearEquation(y3 - y4, x4 - x3, y3*x4 - y4*x3); // Check if the two lines are parallel if (eq1.isParallel(eq2)) { System.out.println("The two lines are parallel"); } else { // Compute the intersecting point double x = eq1.getX(eq2); double y = eq1.getY(eq2); System.out.printf("The intersecting point is (%.2f, %.2f)\n", x, y); } } } class LinearEquation { private double a, b, c; public LinearEquation(double a, double b, double c) { this.a = a; this.b = b; this.c = c; } public double getA() { return a; } public double getB() { return b; } public double getC() { return c; } public boolean isParallel(LinearEquation other) { return Math.abs(a * other.b - b * other.a) < 1e-6; } public double getX(LinearEquation other) { return (c * other.b - b * other.c) / (a * other.b - b * other.a); } public double getY(LinearEquation other) { return (a * other.c - c * other.a) / (a * other.b - b * other.a); } } ``` The program first prompts the user to enter the four endpoints, then creates two `LinearEquation` objects for the two line segments, and finally checks if the two lines are parallel or computes the intersecting point using the `getX` and `getY` methods of the `LinearEquation` class. The `isParallel` method checks if the two lines are parallel by comparing the slopes of the two lines, and the `getX` and `getY` methods solve the system of two linear equations for the intersecting point.
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