第三章第二十五題（幾何：交點）(Geometry: intersecting point)

• *3.25（幾何：交點）兩條直線的交點可以通過下面的線性方程組求解：
  

  這個線性方程組可以應用Cramer法則求解（見程式設計練習題3.3）。如果方程無解，則兩條直線平行。

  編寫一個程式，提示使用者輸入這四個點，然後顯示它們的交點。

  下面是這個程式的執行示例：

  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

• 參考程式碼：

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();
}
}

• 結果顯示：

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