You are given two coordinates (x1, y1) and (x2, y2) of a two-dimensional graph. Find the distance between them.
Examples:
Input : x1, y1 = (3, 4)
x2, y2 = (7, 7)
Output : 5
Input : x1, y1 = (3, 4)
x2, y2 = (4, 3)
Output : 1.41421
Calculate the distance between two points.
We will use the distance formula derived from Pythagorean theorem. The formula for distance between two point (x1, y1) and (x2, y2) is
Distance = 
We can get above formula by simply applying Pythagoras theorem

calculate distance between two points
Below is the implementation of above idea.
Method 1: Without using the inbuilt library,
Python3
def distance(x1, y1, x2, y2):
return (((x2 - x1) * * 2 + (y2 - y1) * * 2 ) * * 0.5 )
print ( distance( 3 , 4 , 4 , 3 ))
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 2: Using the inbuilt library,
C++
#include <bits/stdc++.h>
using namespace std;
float distance( int x1, int y1, int x2, int y2)
{
return sqrt ( pow (x2 - x1, 2) + pow (y2 - y1, 2) * 1.0);
}
int main()
{
cout << distance(3, 4, 4, 3);
return 0;
}
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C
#include <math.h>
#include <stdio.h>
float distance( int x1, int y1, int x2, int y2)
{
return sqrt ( pow (x2 - x1, 2) + pow (y2 - y1, 2) * 1.0);
}
int main()
{
printf ( "%f" , distance(3, 4, 4, 3));
return 0;
}
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Java
class GFG {
static double distance( int x1, int y1, int x2, int y2)
{
return Math.sqrt(Math.pow(x2 - x1, 2 )
+ Math.pow(y2 - y1, 2 ) * 1.0 );
}
public static void main(String[] args)
{
System.out.println(
Math.round(distance( 3 , 4 , 4 , 3 ) * 100000.0 )
/ 100000.0 );
}
}
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Python3
import math
def distance(x1 , y1 , x2 , y2):
return math.sqrt(math. pow (x2 - x1, 2 ) +
math. pow (y2 - y1, 2 ) * 1.0 )
print ( "%.6f" % distance( 3 , 4 , 4 , 3 ))
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C#
using System;
class GFG
{
static double distance( int x1, int y1, int x2, int y2)
{
return Math.Sqrt(Math.Pow(x2 - x1, 2) +
Math.Pow(y2 - y1, 2) * 1.0);
}
public static void Main ()
{
Console.WriteLine(Math.Round(distance(3, 4, 4, 3)
* 100000.0)/100000.0);
}
}
|
PHP
<?php
function distance( $x1 , $y1 , $x2 , $y2 )
{
return sqrt(pow( $x2 - $x1 , 2) +
pow( $y2 - $y1 , 2) * 1.0);
}
echo (distance(3, 4, 4, 3));
?>
|
Javascript
<script>
function distance(x1, y1, x2, y2)
{
return Math.sqrt(Math.pow(x2 - x1, 2) +
Math.pow(y2 - y1, 2) * 1.0);
}
document.write(distance(3, 4, 4, 3));
</script>
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Time Complexity: O(1)
Auxiliary Space: O(1)