Chapter 3 Exercise 25, Introduction to Java Programming, Tenth Edition Y. Daniel LiangY.

*3.25 (Geometry: intersecting point) Two points on line 1 are given as (x1, y1) and (x2, y2) and on line 2 as (x3, y3) and (x4, y4), as shown in Figure 3.8a–b. The intersecting point of the two lines can be found by solving the following linear equation: (y1 - y2)x - (x1 - x2)y = (y1 - y2)x1 - (x1 - x2)y1 (y3 - y4)x - (x3 - x4)y = (y3 - y4)x3 - (x3 - x4)y3 This linear equation can be solved using Cramer’s rule (see Programming Exer- cise 3.3). If the equation has no solutions, the two lines are parallel (Figure 3.8c). Write a program that prompts the user to enter four points and displays the inter- secting point. Here are sample runs:

Enter x1, y1, x2, y2, x3, y3, x4, y4: 2 2 5 -1.0 4.0 2.0 -1.0 -2.0
The intersecting point is at (2.88889, 1.1111)

Enter x1, y1, x2, y2, x3, y3, x4, y4: 2 2 7 6.0 4.0 2.0 -1.0 -2.0
The two lines are parallel 

import java.util.Scanner;
 
public class ProgrammingEx3_25 {
 public static void main(String[] args) {
  System.out.print("Enter x1, y1, x2, y2, x3, y3, x4, y4:");
  Scanner input = new Scanner(System.in);
 
  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();
 
  double a = y1 - y2;
  double b = -(x1 - x2);
  double e = (y1 - y2) * x1 - (x1 - x2) * y1;
  double c = (y3 - y4);
  double d = -(x3 - x4);
  double f = (y3 - y4) * x3 - (x3 - x4) * y3;
 
  if (a * d - b * c == 0) {
   System.out.println("The two lines are parallel.");
   System.exit(0);
  }
 
  double x = (e * d - b * f) / (a * d - b * c);
  double y = (a * f - e * c) / (a * d - b * c);
 
  System.out.print("The intersecting point is at (" + x + ", " + y + ")");
 }
}