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

• Difficulty Level : Easy
• Last Updated : 22 Oct, 2021

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 ``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

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## Javascript

 ``
Output:
`Yes`

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