Given the intercepts of a straight line on both the axes as **m** and **n**. The task is to find the length of the normal on this straight line from the origin.**Examples:**

Input:m = 5, n = 3Output:2.57248Input:m = 13, n = 9Output:7.39973

**Approach:** A normal to a line is a line segment drawn from a point perpendicular to the given line.

Let **p** be the length of the normal drawn from the origin to a line, which subtends an angle **Θ** with the positive direction of x-axis as follows.

Then, we have **cos Θ = p / m** and **sin Θ = p / n**

Since, **sin ^{2} Θ + cos^{2} Θ = 1**

So,

**(p / m)**

^{2}+ (p / n)^{2}= 1We get,

**p = m * n / √(m**

^{2}+ n^{2})Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the normal` `// of the straight line` `float` `normal(` `float` `m, ` `float` `n)` `{` ` ` `// Length of the normal` ` ` `float` `N = (fabsf(m) * fabsf(n))` ` ` `/ ` `sqrt` `((fabsf(m) * fabsf(m))` ` ` `+ (fabsf(n) * fabsf(n)));` ` ` `return` `N;` `}` `// Driver code` `int` `main()` `{` ` ` `float` `m = -5, n = 3;` ` ` `cout << normal(m, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` `// Function to find the normal` `// of the straight line` `static` `float` `normal(` `float` `m, ` `float` `n)` `{` ` ` `// Length of the normal` ` ` `float` `N = (` `float` `) ((Math.abs(m) * Math.abs(n))` ` ` `/ Math.sqrt((Math.abs(m) * Math.abs(m))` ` ` `+ (Math.abs(n) * Math.abs(n))));` ` ` `return` `N;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `float` `m = -` `5` `, n = ` `3` `;` ` ` `System.out.println(normal(m, n));` `}` `}` `// This code has been contributed by 29AjayKumar` |

## Python3

`# Python3 implementation of the approach` `import` `math;` `# Function to find the normal` `# of the straight line` `def` `normal(m, n):` ` ` `# Length of the normal` ` ` `N ` `=` `((` `abs` `(m) ` `*` `abs` `(n)) ` `/` ` ` `math.sqrt((` `abs` `(m) ` `*` `abs` `(m)) ` `+` ` ` `(` `abs` `(n) ` `*` `abs` `(n))));` ` ` `return` `N;` `# Driver code` `m ` `=` `-` `5` `; n ` `=` `3` `;` `print` `(normal(m, n));` `# This code is contributed` `# by Akanksha Rai` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` `// Function to find the normal` `// of the straight line` `static` `float` `normal(` `float` `m, ` `float` `n)` `{` ` ` `// Length of the normal` ` ` `float` `N = (` `float` `)((Math.Abs(m) * Math.Abs(n)) /` ` ` `Math.Sqrt((Math.Abs(m) * Math.Abs(m)) +` ` ` `(Math.Abs(n) * Math.Abs(n))));` ` ` `return` `N;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `float` `m = -5, n = 3;` ` ` `Console.Write(normal(m, n));` `}` `}` `// This code is contributed by Akanksha Rai` |

## PHP

`<?php` `// PHP implementation of the approach` `// Function to find the normal` `// of the straight line` `function` `normal(` `$m` `, ` `$n` `)` `{` ` ` `// Length of the normal` ` ` `$N` `= (` `abs` `(` `$m` `) * ` `abs` `(` `$n` `)) /` ` ` `sqrt((` `abs` `(` `$m` `) * ` `abs` `(` `$m` `)) +` ` ` `(` `abs` `(` `$n` `) * ` `abs` `(` `$n` `)));` ` ` `return` `$N` `;` `}` `// Driver code` `$m` `= -5; ` `$n` `= 3;` `echo` `normal(` `$m` `, ` `$n` `);` `// This code is contributed by Ryuga` `?>` |

## Javascript

`<script>` `// javascript implementation of the approach` `// Function to find the normal` `// of the straight line` `function` `normal(m , n)` `{` ` ` `// Length of the normal` ` ` `var` `N = ((Math.abs(m) * Math.abs(n))` ` ` `/ Math.sqrt((Math.abs(m) * Math.abs(m))` ` ` `+ (Math.abs(n) * Math.abs(n))));` ` ` `return` `N;` `}` `// Driver code` `var` `m = -5, n = 3;` `document.write(normal(m, n).toFixed(5));` `// This code is contributed by Amit Katiyar` `</script>` |

**Output:**

2.57248

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