(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:

/*
(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 Exercise
3.3). If the equation has no solutions, the two lines are parallel Write a program 
that prompts the user to enter four points and displays the intersecting point.
*/
import java.util.Scanner;

public class Exercise_03_25 {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);	// Create Scanner object

		// Prompt the user to enter four points 
		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();

		// Calculate the intersecting point
		// Get a, b, c, d, e, f
		double a = y1 - y2;
		double b = -1 * (x1 - x2);
		double c = y3 - y4;
		double d = -1 * (x3 - x4);
		double e = (y1 - y2) * x1 - (x1 - x2) * y1;
		double f = (y3 - y4) * x3 - (x3 - x4) * y3;

		// Display results
		if (a * d - b * c == 0)
		{
			System.out.println("The two lines are parallel");
		}
		else
		{
			double x = (e * d - b * f) / (a * d - b * c);
			double y = (a * f - e * c) / (a * d - b * c);
			System.out.println("The intersecting point is at (" + x + ", " + y + ")");
		}
	}
}