Check if a right-angled triangle can be formed by the given coordinates
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> |
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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