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# Maximum points of intersection n lines

• Last Updated : 23 Apr, 2021

You are given n straight lines. You have to find a maximum number of points of intersection with these n lines.
Examples:

```Input : n = 4
Output : 6

Input : n = 2
Output :1``` Approach :
As we have n number of line, and we have to find the maximum point of intersection using this n line. So this can be done using the combination. This problem can be thought of as a number of ways to select any two lines among n line. As every line intersects with others that are selected.
So, the total number of points = nC2
Below is the implementation of the above approach:

## C++

 `// CPP program to find maximum intersecting``// points``#include ``using` `namespace` `std;``#define ll long int`  `// nC2 = (n)*(n-1)/2;``ll countMaxIntersect(ll n)``{``   ``return` `(n) * (n - 1) / 2;``}` `// Driver code``int` `main()``{``    ``// n is number of line``    ``ll n = 8;``    ``cout << countMaxIntersect(n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find maximum intersecting``// points` `public` `class` `GFG {``    ` `    ``// nC2 = (n)*(n-1)/2;``    ``static` `long` `countMaxIntersect(``long` `n)``    ``{``       ``return` `(n) * (n - ``1``) / ``2``;``    ``}` `    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``// n is number of line``        ``long` `n = ``8``;``        ``System.out.println(countMaxIntersect(n));`  `    ``}``    ``// This code is contributed by ANKITRAI1``}`

## Python3

 `# Python3 program to find maximum``# intersecting points` `#nC2 = (n)*(n-1)/2``def` `countMaxIntersect(n):``    ``return` `int``(n``*``(n ``-` `1``)``/``2``)` `#Driver code``if` `__name__``=``=``'__main__'``:``    ` `# n is number of line``    ``n ``=` `8``    ``print``(countMaxIntersect(n))` `# this code is contributed by``# Shashank_Sharma`

## C#

 `// C# program to find maximum intersecting``// points``using` `System;` `class` `GFG``{``    ` `    ``// nC2 = (n)*(n-1)/2;``    ``public` `static` `long` `countMaxIntersect(``long` `n)``    ``{``    ``return` `(n) * (n - 1) / 2;``    ``}` `    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``// n is number of line``        ``long` `n = 8;``        ``Console.WriteLine(countMaxIntersect(n));``    ``}``}``// This code is contributed by Soumik`

## PHP

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## Javascript

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Output:
`28`

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