Check if a right-angled triangle can be formed by the given coordinates
Last Updated :
21 Nov, 2022
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++
#include <bits/stdc++.h>
using namespace std;
void checkRightAngled( int X1, int Y1,
int X2, int Y2,
int X3, int Y3)
{
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);
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" ;
}
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
import java.util.*;
class GFG{
static void checkRightAngled( int X1, int Y1,
int X2, int Y2,
int X3, int Y3)
{
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 );
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" );
}
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);
}
}
|
Python3
def checkRightAngled(X1, Y1, X2,
Y2, X3, Y3):
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 )))
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" )
if __name__ = = '__main__' :
X1 = 0 ; X2 = 0 ; X3 = 9 ;
Y1 = 2 ; Y2 = 14 ; Y3 = 2 ;
checkRightAngled(X1, Y1, X2,
Y2, X3, Y3)
|
C#
using System;
class GFG{
static void checkRightAngled( int X1, int Y1,
int X2, int Y2,
int X3, int Y3)
{
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);
if ((A > 0 && B > 0 && C > 0) &&
(A == (B + C) || B == (A + C) ||
C == (A + B)))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
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);
}
}
|
Javascript
<script>
function checkRightAngled
(X1, Y1, X2, Y2, X3, Y3)
{
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);
if (
A > 0 &&
B > 0 &&
C > 0 &&
(A === B + C || B === A + C ||
C === A + B)
)
document.write( "Yes" );
else document.write( "No" );
}
var X1 = 0,
Y1 = 2;
var X2 = 0,
Y2 = 14;
var X3 = 9,
Y3 = 2;
checkRightAngled(X1, Y1, X2, Y2, X3, Y3);
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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