检查线是否穿过原点

给定线的两个坐标分别为(x1，y1)和(x2，y2)，请确定通过这些点的线是否也通过原点。

例子：

Input : (x1, y1) = (10, 0)
        (x2, y2) = (20, 0)
Output : Yes
The line passing through these points
clearly passes through the origin as 
the line is x axis.

Input : (x1, y1) = (1, 28)
        (x2, y2) = (2, 56)
Output : Yes

方法：通过两个点(x1，y1)和(x2，y2)的线的方程式为
y-y1 =((y2-y1)/(x2-x1))(x-x1)+ c
如果线也经过原点，则c = 0，因此线方程变为
y-y1 =((y2-y1)/(x2-x1))(x-x1)
在上面的方程中，保持x = 0，y = 0，
x1(y2-y1)= y1(x2-x1)
因此，如果任何穿过两个坐标(x1，y1)和(x2，y2)的线也穿过原点(0，0)，则必须满足上述方程式。

C++

/* C++ program to find if line passing through
   two coordinates also passes through origin
   or not */
#include 
using namespace std;
 
bool checkOrigin(int x1, int y1, int x2, int y2)
{
   return (x1 * (y2 - y1) == y1 * (x2 - x1));
}
 
// Driver code
int main()
{
    if (checkOrigin(1, 28, 2, 56) == true)
      cout << "Yes";
    else
      cout << "No";
    return 0;
}

Java

// Java program to find if line passing through
// two coordinates also passes through origin
// or not
import java.io.*;
 
class GFG {
     
    static boolean checkOrigin(int x1, int y1,
                                       int x2, int y2)
    {
        return (x1 * (y2 - y1) == y1 * (x2 - x1));
    }
 
    // Driver code
    public static void main (String[] args)
    {
        if (checkOrigin(1, 28, 2, 56) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Ajit.

Python3

# Python program to find if line
# passing through two coordinates
# also passes through origin or not
 
def checkOrigin(x1, y1, x2, y2):
    return (x1 * (y2 - y1) == y1 * (x2 - x1))
     
# Driver code
if (checkOrigin(1, 28, 2, 56) == True):
    print("Yes")
else:
    print("No")
     
# This code is contributed
# by Anant Agarwal.

C#

// C# program to find if line passing through
// two coordinates also passes through origin
// or not
using System;
 
class GFG {
 
    static bool checkOrigin(int x1, int y1,
                            int x2, int y2)
    {
        return (x1 * (y2 - y1) == y1 * (x2 - x1));
    }
 
    // Driver code
    public static void Main()
    {
        if (checkOrigin(1, 28, 2, 56) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to find if line passing through
// two coordinates also passes through origin
// or not
 
    function checkOrigin(x1, y1, x2, y2)
    {
        return (x1 * (y2 - y1) == y1 * (x2 - x1));
    }
  
 
// Driver Code
 
        if (checkOrigin(1, 28, 2, 56) == true)
            document.write("Yes");
        else
            document.write("No");
            
           // This code is contributed by chinmoy1997pal.

输出：

Yes