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++

// 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";
}

Java

// 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

Python3

# 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

C#

// 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

PHP

<?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
?>

Javascript

<script>

// Javascript 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
var m = 3, c = 2;
var x = 1, y = 5;

if (pointIsOnLine(m, c, x, y))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by Rajput-Ji

</script>

Output:

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)