*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
下面是参考答案代码:
import java.util.*;
public class IntersectingPointQuestion25 {
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();
}
}
运行效果：
注：编写程序要养成良好习惯
如：1.文件名要用英文，具体一点
2.注释要英文
3.变量命名要具体，不要抽象（如:a,b,c等等），形式要驼峰化
4.整体书写风格要统一（不要这里是驼峰，那里是下划线，这里的逻辑段落空三行，那里相同的逻辑段落空5行等等）
