Given are two parallel straight lines with slope m, and different y-intercepts b1 & b2.The task is to find the distance between these two parallel lines.
Examples:
Input: m = 2, b1 = 4, b2 = 3 Output: 0.333333 Input: m = -4, b1 = 11, b2 = 23 Output: 0.8
Approach:
- Let PQ and RS be the parallel lines, with equations
y = mx + b1
y = mx + b2
- The distance between these two lines is the distance between the two intersection points of these lines with the perpendicular line.Let that distance be d.
- So, equation of the line perpendicular to PQ and RS can be
y = -x/m
- Now, solving the perpendicular line with PQ and RS separately to get the intersecting points (x1, y1) & (x2, y2), we get,
- From PQ,
y = mx + b1
y = -x/m
(x1, y1) = ( -b1*m/(m^2 + 1), b1/(m^2 + 1))
- From RS,
y = mx + b2
y = -x/m
(x2, y2) = ( -b2*m/(m^2 + 1), b2/(m^2 + 1))
- So, d = distance between (x1, y1) and (x2, y2)
Below is the implementation of the above approach:
C++
// C++ program find the distance// between two parallel lines#include <bits/stdc++.h>using namespace std;// Function to find the distance // between parallel lines double dist(double m, double b1, double b2){ double d = fabs(b2 - b1) / ((m * m) - 1); return d;}// Driver Codeint main(){ double m = 2, b1 = 4, b2 = 3; cout << dist(m, b1, b2); return 0;} |
Java
// Java program find the distance// between two parallel linesclass GFG{ // Function to find the distance // between parallel lines static double dist(double m, double b1, double b2){ double d = Math.abs(b2 - b1) / ((m * m) - 1); return d;}// Driver Codepublic static void main(String[] args){ double m = 2, b1 = 4, b2 = 3; System.out.println(dist(m, b1, b2));}}// This code is contributed by Code_Mech. |
Python3
# Python3 program find the distance# between two parallel lines# Function to find the distance # between parallel lines def dist(m, b1, b2): d = abs(b2 - b1) / ((m * m) - 1); return d;# Driver Codedef main(): m, b1, b2 =2,4, 3; print(dist(m, b1, b2));if __name__ == '__main__': main()# This code contributed by PrinciRaj1992 |
C#
// C# program find the distance// between two parallel linesusing System;class GFG{ // Function to find the distance // between parallel lines static double dist(double m, double b1, double b2){ double d = Math.Abs(b2 - b1) / ((m * m) - 1); return d;}// Driver Codepublic static void Main(){ double m = 2, b1 = 4, b2 = 3; Console.Write(dist(m, b1, b2));}}// This code is contributed by Akanksha Rai |
PHP
<?php// PHP program find the distance // between two parallel lines // Function to find the distance // between parallel linesfunction dist($m, $b1, $b2) { $d = abs($b2 - $b1) / (($m * $m) - 1); return $d; } // Driver Code $m = 2;$b1 = 4;$b2 = 3; echo dist($m, $b1, $b2);// This code is contributed by Ryuga?> |
Javascript
<script>// javascript program find the distance// between two parallel lines // Function to find the distance // between parallel lines function dist(m, b1 , b2){ var d = Math.abs(b2 - b1) / ((m * m) - 1); return d;}// Driver Codevar m = 2, b1 = 4, b2 = 3;document.write(dist(m, b1, b2).toFixed(5));// This code contributed by Princi Singh </script> |
Output:
0.333333
Time Complexity: O(1)
Auxiliary Space: O(1)
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