*3.25（几何：交点）两条直线的交点可以通过下面的线性方程组求解：
这个线性方程组可以应用Cramer法则求解（见编程练习题3.3）。如果方程无解，则两条直线平行。
编写一个程序，提示用户输入这四个点，然后显示它们的交点。
下面是这个程序的运行示例：
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
*3.25(Geometry: intersecting point) The intersecting point of the two lines can be found by solving the following linear equations: 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.
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
参考代码：
package chapter03; import java.util.Scanner; public class Code_25 {public static void main(String[] args) { double x1, y1, x2, y2, x3, y3, x4, y4; double a, b, c, d, e, f, x, y, discriminant; System.out.print("Enter x1, y1, x2, y2, x3, y3, x4, y4: "); Scanner input = new Scanner(System.in); x1 = input.nextDouble(); y1 = input.nextDouble(); x2 = input.nextDouble(); y2 = input.nextDouble(); x3 = input.nextDouble(); y3 = input.nextDouble(); x4 = input.nextDouble(); y4 = input.nextDouble(); a = y1 - y2; b = x2 - x1; c = y3 - y4; d = x4 - x3; e = a * x1 + b * y1; f = c * x3 + d * y3; discriminant = a * d - b * c; if(discriminant != 0) { x = (e * d - b * f) / discriminant; y = (a * f - e * c) / discriminant; System.out.println("The intersecting point is at(" + x + ", " + y + ")"); } else System.out.println("The two lines are parallel"); input.close(); } } 结果显示： 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(2.888888888888889, 1.1111111111111112) Process finished with exit code 0