Count number of pairs of lines intersecting at a Point

Given N lines are in the form a*x + b*y = c (a>0 or a==0 & b>0). Find the number of pairs of lines intersecting at a point.


Input: N=5
x + y = 2
x + y = 4
x = 1
x – y = 2
y = 3
Output: 9

Input: N=2
x + 2y = 2
x + 2y = 4
Output: 0


  • Parallel lines never intersect so a method is needed to exclude parallel lines for each line.
  • The slope of a line can be represented as pair(a, b). Construct a map with key as slope and value as a set with c as entries in it so that it has an account of the parallel lines.
  • Iterate over the lines add them to the map and maintain a variable Tot which counts the total number of lines till now.
  • Now for each line update the Tot variable then add Tot to the answer and subtract the number of parallel lines to that line including itself.

Below is the implementation of the above approach:





// C++ implementation to calculate
// pair of intersecting lines
#include <bits/stdc++.h>
using namespace std;
// Function to return the number 
// of intersecting pair of lines
void numberOfPairs(int a[],int b[],int c[],int N){
    int count = 0, Tot = 0;
    // Construct a map of slope and 
    // corresponding c value
    map<pair<int, int>, set<int> > LineMap;
    // iterate over each line
    for (int i = 0; i < N; i++) {
        // Slope can be represented
        // as pair(a, b)
        pair<int, int> Slope =
                     make_pair(a[i], b[i]);
        // Checking if the line does 
        // not already exist
        if (!LineMap[Slope].count(c[i])){
            // maintaining a count
            // of total lines
            // subtracting the count of
            // parallel lines including itself
            count += Tot - 
    cout << count << endl;
// Driver code
int main()
    // A line can be represented as ax+by=c
    // such  that (a>0 || (a==0 & b>0) ) 
    // a and b are  already in there lowest
    // form i.e gcd(a, b)=1
    int N = 5;
    int a[] = { 1, 1, 1, 1, 0 };
    int b[] = { 1, 1, 0, -1, 1 };
    int c[] = { 2, 4, 1, 2, 3 };
    return 0;




Time Complexity: O(N*log(N))

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