# Maximum distinct lines passing through a single point

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


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

• 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  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 s=new HashSet();         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 s = new HashSet();         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

 

Output:

2


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