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> |
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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