Check if a right-angled triangle can be formed by the given coordinates - GeeksforGeeks

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++
Java
Python3
C#
Javascript
// 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 coordinatesvoid 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 Codeint 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 program for the above approachimport java.util.*; class GFG{     // Function to check if right-angled// triangle can be formed by the// given coordinatesstatic 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 Codepublic 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 program for the# above approach # Function to check if right-angled# triangle can be formed by the# given coordinatesdef 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 codeif __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# program for the above approachusing System; class GFG{     // Function to check if right-angled// triangle can be formed by the// given coordinatesstatic 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 Codepublic 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
<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>
Yes
Time Complexity: O(logN) Auxiliary Space: O(1)
virusbuddha
rutvik_56
Rohit_ranjan
rdtank
triangle
Geometric
Mathematical
School Programming
Mathematical
Geometric
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