# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// CPP program to find point that divides ` `// given line in given ratio. ` `#include ` `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; ` `} `

## Java

 `// 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); ` `    ``} ` `} `

## Python

 `# 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) `

## C#

 `// 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. `

## PHP

 ` `

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.```

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