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 = 3
Output: 2.57248
Input: m = 13, n = 9
Output: 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 ? with the positive direction of x-axis as follows.
Then, we have cos ? = p / m and sin ? = p / n
Since, sin2 ? + cos2 ? = 1
So, (p / m)2 + (p / n)2 = 1
We get, p = m * n / ?(m2 + n2)
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 linefloat 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 codeint main(){ float m = -5, n = 3; cout << normal(m, n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to find the normal// of the straight linestatic 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 codepublic 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 approachimport math;# Function to find the normal# of the straight linedef 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 codem = -5; n = 3;print(normal(m, n));# This code is contributed# by Akanksha Rai |
C#
// C# implementation of the approachusing System;class GFG{// Function to find the normal// of the straight linestatic 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 codepublic 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 linefunction 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 codevar m = -5, n = 3;document.write(normal(m, n).toFixed(5));// This code is contributed by Amit Katiyar </script> |
Output:
2.57248
Time Complexity: O(1)
Auxiliary Space: O(1)
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