Given lines represented by two points and . The task is to find maximum number of lines which can pass through a single point, without superimposing (or covering) any other line. We can move any line but not rotate it.

**Examples:**

Input : Line 1 : x1 = 1, y1 = 1, x2 = 2, y2 = 2 Line 2 : x2 = 2, y1 = 2, x2 = 4, y2 = 10 Output : 2 There are two lines. These two lines are not parallel, so both of them will pass through a single point. Input : Line 1 : x1 = 1, y1 = 5, x2 = 1, y2 = 10 Line 2 : x2 = 5, y1 = 1, x2 = 10, y2 = 1 Output : 2

- Represent lines as pair where line can be given as , called line slope form. We can now see that we can change the
**c**for any line, but cannot modify**m**. - Lines having same value of
**m**parallel, given that**(c1 ≠ c2)**. Also no two parallel lines can pass through same point without superimposing to each other. - So, our problem reduces to finding different values of slopes from given set of lines.

We can calculate slope of a line as , add them to a set and count the number of distinct values of slope in set. But we have to handle vertical lines separately.

So, if then, **slope = INT_MAX**.

Otherwise, **slope = **.

Below is the implementation of the approach.

## C++

`// C++ program to find maximum number of lines ` `// which can pass through a single point ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to find maximum lines which passes ` `// through a single point ` `int` `maxLines(` `int` `n, ` `int` `x1[], ` `int` `y1[], ` ` ` `int` `x2[], ` `int` `y2[]) ` `{ ` ` ` `unordered_set<` `double` `> s; ` ` ` ` ` `double` `slope; ` ` ` `for` `(` `int` `i = 0; i < n; ++i) { ` ` ` `if` `(x1[i] == x2[i]) ` ` ` `slope = INT_MAX; ` ` ` `else` ` ` `slope = (y2[i] - y1[i]) * 1.0 ` ` ` `/ (x2[i] - x1[i]) * 1.0; ` ` ` ` ` `s.insert(slope); ` ` ` `} ` ` ` ` ` `return` `s.size(); ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 }, ` ` ` `x2[] = { 2, 4 }, y2[] = { 2, 10 }; ` ` ` `cout << maxLines(n, x1, y1, x2, y2); ` ` ` `return` `0; ` `} ` `// This code is written by ` `// Sanjit_Prasad ` |

## Java

`// Java program to find maximum number of lines ` `// which can pass through a single point ` ` ` `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` ` `class` `GFG{ ` ` ` `// function to find maximum lines which passes ` `// through a single point ` `static` `int` `maxLines(` `int` `n, ` `int` `x1[], ` `int` `y1[], ` ` ` `int` `x2[], ` `int` `y2[]) ` `{ ` ` ` `Set<Double> s=` `new` `HashSet<Double>(); ` ` ` ` ` `double` `slope; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; ++i) { ` ` ` `if` `(x1[i] == x2[i]) ` ` ` `slope = Integer.MAX_VALUE; ` ` ` `else` ` ` `slope = (y2[i] - y1[i]) * ` `1.0` ` ` `/ (x2[i] - x1[i]) * ` `1.0` `; ` ` ` ` ` `s.add(slope); ` ` ` `} ` ` ` ` ` `return` `s.size(); ` `} ` ` ` `// Driver program ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `n = ` `2` `, x1[] = { ` `1` `, ` `2` `}, y1[] = { ` `1` `, ` `2` `}, ` ` ` `x2[] = { ` `2` `, ` `4` `}, y2[] = { ` `2` `, ` `10` `}; ` ` ` `System.out.print(maxLines(n, x1, y1, x2, y2)); ` `} ` `} ` `// This code is written by ` `// Subhadeep ` |

## Python3

`# Python3 program to find maximum number ` `# of lines which can pass through a ` `# single point ` `import` `sys ` `# function to find maximum lines ` `# which passes through a single point ` `def` `maxLines(n, x1, y1, x2, y2): ` ` ` ` ` `s ` `=` `[]; ` ` ` ` ` `slope` `=` `sys.maxsize; ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `if` `(x1[i] ` `=` `=` `x2[i]): ` ` ` `slope ` `=` `sys.maxsize; ` ` ` `else` `: ` ` ` `slope ` `=` `(y2[i] ` `-` `y1[i]) ` `*` `1.0` `/` `(x2[i] ` `-` `x1[i]) ` `*` `1.0` `; ` ` ` ` ` `s.append(slope); ` ` ` ` ` `return` `len` `(s); ` ` ` `# Driver Code ` `n ` `=` `2` `; ` `x1 ` `=` `[ ` `1` `, ` `2` `]; ` `y1 ` `=` `[` `1` `, ` `2` `]; ` `x2 ` `=` `[` `2` `, ` `4` `]; ` `y2 ` `=` `[` `2` `, ` `10` `]; ` `print` `(maxLines(n, x1, y1, x2, y2)); ` ` ` `# This code is contributed by mits ` |

## C#

`// C# program to find maximum number of lines ` `// which can pass through a single point ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{ ` ` ` `// function to find maximum lines which passes ` `// through a single point ` `static` `int` `maxLines(` `int` `n, ` `int` `[]x1, ` `int` `[]y1, ` ` ` `int` `[]x2, ` `int` `[]y2) ` `{ ` ` ` `HashSet<Double> s = ` `new` `HashSet<Double>(); ` ` ` ` ` `double` `slope; ` ` ` `for` `(` `int` `i = 0; i < n; ++i) ` ` ` `{ ` ` ` `if` `(x1[i] == x2[i]) ` ` ` `slope = ` `int` `.MaxValue; ` ` ` `else` ` ` `slope = (y2[i] - y1[i]) * 1.0 ` ` ` `/ (x2[i] - x1[i]) * 1.0; ` ` ` ` ` `s.Add(slope); ` ` ` `} ` ` ` ` ` `return` `s.Count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ` `int` `n = 2; ` ` ` `int` `[]x1 = { 1, 2 }; ` `int` `[]y1 = { 1, 2 }; ` ` ` `int` `[]x2 = { 2, 4 }; ` `int` `[]y2 = { 2, 10 }; ` ` ` `Console.Write(maxLines(n, x1, y1, x2, y2)); ` `} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */` |

## PHP

`<?php ` `// PHP program to find maximum number ` `// of lines which can pass through a ` `// single point ` ` ` `// function to find maximum lines ` `// which passes through a single point ` `function` `maxLines(` `$n` `, ` `$x1` `, ` `$y1` `, ` `$x2` `, ` `$y2` `) ` `{ ` ` ` `$s` `= ` `array` `(); ` ` ` ` ` `$slope` `; ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ++` `$i` `) ` ` ` `{ ` ` ` `if` `(` `$x1` `[` `$i` `] == ` `$x2` `[` `$i` `]) ` ` ` `$slope` `= PHP_INT_MAX; ` ` ` `else` ` ` `$slope` `= (` `$y2` `[` `$i` `] - ` `$y1` `[` `$i` `]) * 1.0 / ` ` ` `(` `$x2` `[` `$i` `] - ` `$x1` `[` `$i` `]) * 1.0; ` ` ` ` ` `array_push` `(` `$s` `, ` `$slope` `); ` ` ` `} ` ` ` ` ` `return` `count` `(` `$s` `); ` `} ` ` ` `// Driver Code ` `$n` `= 2; ` `$x1` `= ` `array` `( 1, 2 ); ` `$y1` `= ` `array` `(1, 2); ` `$x2` `= ` `array` `(2, 4); ` `$y2` `= ` `array` `(2, 10); ` `echo` `maxLines(` `$n` `, ` `$x1` `, ` `$y1` `, ` `$x2` `, ` `$y2` `); ` ` ` `// This code is contributed by mits ` `?> ` |

**Output:**

2

**Time Complexity:**

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