# Maximum points of intersection n lines

You are given n straight lines. You have to find maximum number of point 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 maximum point of intersection using these n line. So this can be done using combination. This problem can be think as number of ways to select any two line among n line. As every line intersect with other that is selected.

So, total number of points = nC2

Below is the implementation of above approach:

## C++

`// CPP program to find maximum intersecting ` `// points ` `#include <bits/stdc++.h> ` `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; ` `} ` |

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## 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 ` `} ` |

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## 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 ` |

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## 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 ` |

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## PHP

`<?PHP ` `// PHP program to find maximum intersecting ` `// points ` ` ` `// nC2 = (n)*(n-1)/2; ` `function` `countMaxIntersect(` `$n` `) ` `{ ` ` ` `return` `(` `$n` `) * (` `$n` `- 1) / 2; ` `} ` ` ` `// Driver code ` ` ` `// n is number of line ` `$n` `= 8; ` `echo` `countMaxIntersect(` `$n` `) . ` `"\n"` `; ` ` ` `// This code is contributed by ChitraNayal ` `?> ` |

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**Output:**

28

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