检查是否可以通过给定的坐标形成直角三角形

给定三个笛卡尔坐标，任务是检查给定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“是” 。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：
这个想法是用毕达哥拉斯定理检查一个直角三角形是否可行。通过连接给定的坐标来计算三角形的三个边的长度。令边为A，B和C。当且仅当A ² + B ² = C ^2时，给定的三角形才是直角。如果条件成立，则打印“是”。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

输出：

Yes

时间复杂度： O(logN)
辅助空间： O(1)