Distance between two parallel lines
Last Updated :
07 Jun, 2022
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++
#include <bits/stdc++.h>
using namespace std;
double dist(double m, double b1, double b2)
{
double d = fabs(b2 - b1) / ((m * m) - 1);
return d;
}
int main()
{
double m = 2, b1 = 4, b2 = 3;
cout << dist(m, b1, b2);
return 0;
}
|
Java
class GFG
{
static double dist(double m,
double b1, double b2)
{
double d = Math.abs(b2 - b1) /
((m * m) - 1);
return d;
}
public static void main(String[] args)
{
double m = 2, b1 = 4, b2 = 3;
System.out.println(dist(m, b1, b2));
}
}
|
Python3
def dist(m, b1, b2):
d = abs(b2 - b1) / ((m * m) - 1);
return d;
def main():
m, b1, b2 =2,4, 3;
print(dist(m, b1, b2));
if __name__ == '__main__':
main()
|
C#
using System;
class GFG
{
static double dist(double m,
double b1, double b2)
{
double d = Math.Abs(b2 - b1) /
((m * m) - 1);
return d;
}
public static void Main()
{
double m = 2, b1 = 4, b2 = 3;
Console.Write(dist(m, b1, b2));
}
}
|
PHP
<?php
function dist($m, $b1, $b2)
{
$d = abs($b2 - $b1) / (($m * $m) - 1);
return $d;
}
$m = 2;
$b1 = 4;
$b2 = 3;
echo dist($m, $b1, $b2);
?>
|
Javascript
<script>
function dist(m, b1 , b2)
{
var d = Math.abs(b2 - b1) /
((m * m) - 1);
return d;
}
var m = 2, b1 = 4, b2 = 3;
document.write(dist(m, b1, b2).toFixed(5));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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