# Distance between two parallel lines

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

1. Let PQ and RS be the parallel lines, with equations
y = mx + b1
y = mx + b2

2. 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.
3. So, equation of the line perpendicular to PQ and RS can be
y = -x/m
4. Now, solving the perpendicular line with PQ and RS separately to get the intersecting points (x1, y1) & (x2, y2), we get,
5. From PQ,
y = mx + b1
y = -x/m
(x1, y1) = ( -b1*m/(m^2 + 1), b1/(m^2 + 1))
6. From RS,
y = mx + b2
y = -x/m
(x2, y2) = ( -b2*m/(m^2 + 1), b2/(m^2 + 1))
7. 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  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 Code  int main()  {      double m = 2, b1 = 4, b2 = 3;      cout << dist(m, b1, b2);      return 0;  }

## Java

 // Java program find the distance  // between two parallel lines  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 Code  public 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 Code  def 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 lines  using 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 Code  public static void Main()  {      double m = 2, b1 = 4, b2 = 3;      Console.Write(dist(m, b1, b2));  }  }     // This code is contributed by Akanksha Rai

## PHP

 

Output:

0.333333


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