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Check if a right-angled triangle can be formed by the given coordinates
  • Difficulty Level : Hard
  • Last Updated : 30 Apr, 2021

Given three Cartesian coordinates, the task is to check if a right-angled triangle can be formed by the given coordinates. If it is possible to create a right-angled triangle, print Yes. Otherwise, print No.
Examples: 
 

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 
 

 

Approach: 
The idea is to use the Pythagoras Theorem to check if a right-angled triangle is possible or not. Calculate the length of the three sides of the triangle by joining the given coordinates. Let the sides be A, B, and C. The given triangle is right-angled if and only if A2 + B2 = C2. Print Yes if the condition holds true. Otherwise, print No.

Below is the implementation of the above approach:
 

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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




<script>
 
      // 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);
       
</script>
Output: 
Yes

Time Complexity: O(logN) 
Auxiliary Space: O(1)

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