# 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++

`// 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;` `}` |

## 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

`<?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.` `?>` |

## Javascript

`<script>` ` ` `// JavaScript program to find point that divides` `// given line in given ratio` ` ` ` ` `function` `section(x1, x2, y1, y2, m, n)` ` ` `{` ` ` `// Applying section formula` ` ` `let x = ((n * x1) + (m * x2)) /` ` ` `(m + n);` ` ` `let y = ((n * y1) + (m * y2)) /` ` ` `(m + n);` ` ` ` ` ` ` `// Printing result` ` ` `document.write(` `"("` `+ x + ` `", "` `+ y + ` `")"` `);` ` ` `}` ` ` `// Driver Code` ` ` ` ` `let x1 = 2, x2 = 4, y1 = 4,` ` ` `y2 = 6, m = 2, n = 3;` ` ` `section(x1, x2, y1, y2, m, n)` ` ` ` ` `// This code is contributed by avijitmondal1998.` `</script>` |

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