Check whether the point (x, y) lies on a given line

Given the values of m and c for the equation of a line y = (m * x) + c, the task is to find whether the point (x, y) lies on the given line.

Examples:

Input: m = 3, c = 2, x = 1, y = 5
Output: Yes
m * x + c = 3 * 1 + 2 = 3 + 2 = 5 which is equal to y
Hence, the given point satisfies the line’s equation

Input: m = 5, c = 2, x = 2, y = 5
Output: No

Approach: In order for the given point to lie on the line, it must satisfy the equation of the line. Check whether y = (m * x) + c holds true.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that return true if
// the given point lies on the given line
bool pointIsOnLine(int m, int c, int x, int y)
{
    // If (x, y) satisfies the equation of the line
    if (y == ((m * x) + c))
        return true;
  
    return false;
}
  
// Driver code
int main()
{
    int m = 3, c = 2;
    int x = 1, y = 5;
  
    if (pointIsOnLine(m, c, x, y))
        cout << "Yes";
    else
        cout << "No";
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
  
class GFG
{
  
// Function that return true if
// the given point lies on the given line
static boolean pointIsOnLine(int m, int c,
                        int x, int y)
{
    // If (x, y) satisfies the equation 
    // of the line
    if (y == ((m * x) + c))
        return true;
  
    return false;
}
  
// Driver code
public static void main(String[] args)
{
    int m = 3, c = 2;
    int x = 1, y = 5;
  
    if (pointIsOnLine(m, c, x, y))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code has been contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function that return true if the 
# given point lies on the given line 
def pointIsOnLine(m, c, x, y):
      
    # If (x, y) satisfies the 
    # equation of the line 
    if (y == ((m * x) + c)): 
        return True
  
    return False
  
# Driver code 
m = 3; c = 2
x = 1; y = 5
  
if (pointIsOnLine(m, c, x, y)): 
    print("Yes"); 
else:
    print("No"); 
      
# This code is contributed by mits

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function that return true if
// the given point lies on the given line
static bool pointIsOnLine(int m, int c,
                          int x, int y)
{
    // If (x, y) satisfies the equation 
    // of the line
    if (y == ((m * x) + c))
        return true;
  
    return false;
}
  
// Driver code
public static void Main()
{
    int m = 3, c = 2;
    int x = 1, y = 5;
  
    if (pointIsOnLine(m, c, x, y))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by Akanksha Rai

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach 
  
// Function that return true if the 
// given point lies on the given line 
function pointIsOnLine($m, $c, $x, $y
      
    // If (x, y) satisfies the equation
    // of the line 
    if ($y == (($m * $x) + $c)) 
        return true; 
  
    return false; 
  
// Driver code 
$m = 3; $c = 2; 
$x = 1; $y = 5; 
  
if (pointIsOnLine($m, $c, $x, $y)) 
    echo "Yes"
else
    echo "No"
      
// This code is contributed by Ryuga
?>

chevron_right


Output:

Yes


My Personal Notes arrow_drop_up

A fallen star which will rise again

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


3


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.