Section formula (Point that divides a line in given ratio)

Given two coordinates (x1, y1) and (x2, y2), and m and n, find the co-ordinates that divides that divides the line joining (x1, y1) and (x2, y2) in the ratio m : n

Examples:

Input : x1 = 1, y1 = 0, x2 = 2 y2 = 5,
        m = 1, n = 1
Output : (1.5, 2.5)
Explanation: co-ordinates (1.5, 2.5)
divides the line in ratio 1 : 1

Input : x1 = 2, y1 = 4, x2 = 4, y2 = 6,
        m = 2, n = 3
Output : (2.8, 4.8)
Explanation: (2.8, 4.8) divides the line
in the ratio 2:3

The section formula tells us the coordinates of the point which divides a given line segment into two parts such that their lengths are in the ratio m : n

C++

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// CPP program to find point that divides
// given line in given ratio.
#include <iostream>
using namespace std;
  
// Function to find the section of the line
void section(double x1, double x2, double y1,
              double y2, double m, double n)
{
    // Applying section formula
    double x = ((n * x1) + (m * x2)) /
                            (m + n);
    double y = ((n * y1) + (m * y2)) /
                             (m + n);
  
    // Printing result
    cout << "(" << x << ", ";
    cout << y << ")" << endl;
}
  
// Driver code
int main()
{
    double x1 = 2, x2 = 4, y1 = 4,
           y2 = 6, m = 2, n = 3;
    section(x1, x2, y1, y2, m, n);
    return 0;
}

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Java

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// Java program to find point that divides
// given line in given ratio.
import java.io.*;
  
class sections {
    static void section(double x1, double x2,
                        double y1, double y2,
                        double m, double n)
    {
        // Applying section formula
        double x = ((n * x1) + (m * x2)) /
                    (m + n);
        double y = ((n * y1) + (m * y2)) /
                    (m + n);
  
  
        // Printing result
        System.out.println("(" + x + ", " + y + ")");
    }
  
    public static void main(String[] args)
    {
        double x1 = 2, x2 = 4, y1 = 4,
               y2 = 6, m = 2, n = 3;
        section(x1, x2, y1, y2, m, n);
    }
}

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Python

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# Python program to find point that divides
# given line in given ratio.
def section(x1, x2, y1, y2, m, n):
  
    # Applying section formula
    x = (float)((n * x1)+(m * x2))/(m + n)
    y = (float)((n * y1)+(m * y2))/(m + n)
  
    # Printing result
    print (x, y)
  
x1 = 2
x2 = 4
y1 = 4
y2 = 6
m = 2
n = 3
section(x1, x2, y1, y2, m, n)

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C#

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// C# program to find point that divides
// given line in given ratio.
using System;
  
class GFG {
      
    static void section(double x1, double x2,
                        double y1, double y2,
                          double m, double n)
    {
          
        // Applying section formula
        double x = ((n * x1) + (m * x2)) /
                                    (m + n);
                                      
        double y = ((n * y1) + (m * y2)) /
                                   (m + n);
  
        // Printing result
        Console.WriteLine("(" + x + ", " + y + ")");
    }
  
    // Driver code
    public static void Main()
    {
          
        double x1 = 2, x2 = 4, y1 = 4,
                y2 = 6, m = 2, n = 3;
                  
        section(x1, x2, y1, y2, m, n);
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find point that 
// divides given line in given ratio.
  
// Function to find the
// section of the line
function section($x1, $x2, $y1,
                 $y2, $m, $n)
{
      
    // Applying section formula
    $x = (($n * $x1) + ($m * $x2))
                     / ($m + $n);
                              
    $y = (($n * $y1) + ($m * $y2)) 
                     / ($m + $n);
  
    // Printing result
    echo("(" . $x . ", ");
    echo($y . ")");
}
  
// Driver code
$x1 = 2; $x2 = 4; $y1 = 4;
$y2 = 6; $m = 2; $n = 3;
section($x1, $x2, $y1, $y2, $m, $n);
      
// This code is contributed by Ajit.
?>

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Output:

(2.8, 4.8)

How does this work?

From our diagram, we can see,
PS = x – x1 and RT = x2 – x

We are given,

PR/QR = m/n

Using similarity, we can write
RS/QT = PS/RT = PR/QR

Therefore, we can write
 PS/RR = m/n
 (x - x1) / (x2 - x) = m/n

From above, we get
  x = (mx2 + nx1) / (m + n)

Similarly, we can solve for y.


References:

http://doubleroot.in/lessons/coordinate-geometry-basics/section-formula/#.WjYXQvbhU8o



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