Check if a line passes through the origin

Given two coordinates of a line as (x1, y1) and (x2, y2), find if the line passing through these points also passes through origin or not.

Examples: 

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

Approach: Equation of a line passing through two points (x1, y1) and (x2, y2) is given by 
y-y1 = ((y2-y1) / (x2-x1))(x-x1) + c 
If line is also passing through origin, then c=0, so equation of line becomes 
y-y1 = ((y2-y1) / (x2-x1))(x-x1) 
Keeping x=0, y=0 in the above equation we get, 
x1(y2-y1) = y1(x2-x1) 
So above equation must be satisfied if any line passing through two coordinates (x1, y1) and (x2, y2) also passes through origin (0, 0).

C++

/* C++ program to find if line passing through
   two coordinates also passes through origin
   or not */
#include <bits/stdc++.h>
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.

PHP

<?php
// PHP 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)
    echo("Yes");
else
    echo("No");

// This code is contributed by Ajit.
?>

Javascript

<script>

// 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.
</script>

Output:

Yes

Time Complexity: O(1) 
Auxiliary Space: O(1)