检查给定的坐标是否可以形成直角三角形

给定三个笛卡尔坐标，任务是检查给定的坐标是否可以形成直角三角形。如果可以创建直角三角形，则打印Yes 。否则，打印No 。
例子：

Input: X1=0, Y1=5, X2=19, Y2=5, X3=0, Y3=0 
Output: Yes 
Explanation: 
Length of side connecting points (X1, Y1) and (X2, Y2) is 12. 
Length of side connecting points (X2, Y2) and (X3, Y3) is 15. 
Length of side connecting points (X1, Y1) and (X3, Y3) is 9. 
12² + 9² = 15². 
Therefore, a right-angled triangle can be made.

Input: X1=5, Y1=14, X2=6, Y2=13, X3=8, Y3=7 
Output: No 
 

方法：
这个想法是使用毕达哥拉斯定理来检查直角三角形是否可能。通过连接给定的坐标来计算三角形三边的长度。设边为A、B和C。给定的三角形是直角当且仅当A ² + B ² = C ² 。如果条件成立，则打印Yes 。否则，打印编号。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if right-angled
// triangle can be formed by the
// given coordinates
void checkRightAngled(int X1, int Y1,
                      int X2, int Y2,
                      int X3, int Y3)
{
    // Calculate the sides
    int A = (int)pow((X2 - X1), 2)
            + (int)pow((Y2 - Y1), 2);
 
    int B = (int)pow((X3 - X2), 2)
            + (int)pow((Y3 - Y2), 2);
 
    int C = (int)pow((X3 - X1), 2)
            + (int)pow((Y3 - Y1), 2);
 
    // Check Pythagoras Formula
    if ((A > 0 and B > 0 and C > 0)
        and (A == (B + C) or B == (A + C)
             or C == (A + B)))
        cout << "Yes";
 
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    int X1 = 0, Y1 = 2;
    int X2 = 0, Y2 = 14;
    int X3 = 9, Y3 = 2;
 
    checkRightAngled(X1, Y1, X2,
                     Y2, X3, Y3);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check if right-angled
// triangle can be formed by the
// given coordinates
static void checkRightAngled(int X1, int Y1,
                             int X2, int Y2,
                             int X3, int Y3)
{
     
    // Calculate the sides
    int A = (int)Math.pow((X2 - X1), 2) +
            (int)Math.pow((Y2 - Y1), 2);
 
    int B = (int)Math.pow((X3 - X2), 2) +
            (int)Math.pow((Y3 - Y2), 2);
 
    int C = (int)Math.pow((X3 - X1), 2) +
            (int)Math.pow((Y3 - Y1), 2);
 
    // Check Pythagoras Formula
    if ((A > 0 && B > 0 && C > 0) &&
        (A == (B + C) || B == (A + C) ||
         C == (A + B)))
        System.out.println("Yes");
    else
        System.out.println("No");
}
 
// Driver Code
public static void main(String s[])
{
    int X1 = 0, Y1 = 2;
    int X2 = 0, Y2 = 14;
    int X3 = 9, Y3 = 2;
     
    checkRightAngled(X1, Y1, X2, Y2, X3, Y3);
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program for the
# above approach
 
# Function to check if right-angled
# triangle can be formed by the
# given coordinates
def checkRightAngled(X1, Y1, X2,
                     Y2, X3, Y3):
     
    # Calculate the sides
    A = (int(pow((X2 - X1), 2)) +
         int(pow((Y2 - Y1), 2)))
    B = (int(pow((X3 - X2), 2)) +
         int(pow((Y3 - Y2), 2)))
    C = (int(pow((X3 - X1), 2)) +
         int(pow((Y3 - Y1), 2)))
     
    # Check Pythagoras Formula
    if ((A > 0 and B > 0 and C > 0) and
        (A == (B + C) or B == (A + C) or
         C == (A + B))):
        print("Yes")
    else:
        print("No")
 
# Driver code
if __name__=='__main__':
     
    X1 = 0; X2 = 0; X3 = 9;
    Y1 = 2; Y2 = 14; Y3 = 2;
     
    checkRightAngled(X1, Y1, X2,
                     Y2, X3, Y3)
     
# This code is contributed by virusbuddah_

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if right-angled
// triangle can be formed by the
// given coordinates
static void checkRightAngled(int X1, int Y1,
                             int X2, int Y2,
                             int X3, int Y3)
{
     
    // Calculate the sides
    int A = (int)Math.Pow((X2 - X1), 2) +
            (int)Math.Pow((Y2 - Y1), 2);
 
    int B = (int)Math.Pow((X3 - X2), 2) +
            (int)Math.Pow((Y3 - Y2), 2);
 
    int C = (int)Math.Pow((X3 - X1), 2) +
            (int)Math.Pow((Y3 - Y1), 2);
 
    // Check Pythagoras Formula
    if ((A > 0 && B > 0 && C > 0) &&
        (A == (B + C) || B == (A + C) ||
         C == (A + B)))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
 
// Driver Code
public static void Main(String []s)
{
    int X1 = 0, Y1 = 2;
    int X2 = 0, Y2 = 14;
    int X3 = 9, Y3 = 2;
     
    checkRightAngled(X1, Y1, X2, Y2, X3, Y3);
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
      // JavaScript program to implement
      // the above approach
      // Function to check if right-angled
      // triangle can be formed by the
      // given coordinates
      function checkRightAngled
      (X1, Y1, X2, Y2, X3, Y3)
      {
        // Calculate the sides
        var A = Math.pow(X2 - X1, 2) +
        Math.pow(Y2 - Y1, 2);
 
        var B = Math.pow(X3 - X2, 2) +
        Math.pow(Y3 - Y2, 2);
 
        var C = Math.pow(X3 - X1, 2) +
        Math.pow(Y3 - Y1, 2);
 
        // Check Pythagoras Formula
        if (
          A > 0 &&
          B > 0 &&
          C > 0 &&
          (A === B + C || B === A + C ||
          C === A + B)
        )
          document.write("Yes");
        else document.write("No");
      }
 
      // Driver Code
      var X1 = 0,
        Y1 = 2;
      var X2 = 0,
        Y2 = 14;
      var X3 = 9,
        Y3 = 2;
      checkRightAngled(X1, Y1, X2, Y2, X3, Y3);
       

输出：

Yes

时间复杂度： O(logN)
辅助空间： O(1)