Maximum points on a line

/*
Logic:
We can find the slope of line from two points by slope=(y2-y1)/(x2-x1).
We will iterate through the points and consider all the pairs of points.
There is one temporary map which will store the slope count for every pairs
for each ith iteration and finally pairs with maximum points of same slope
will be the answer.
There is one catch that when line will be vertical then (x2-x1) ie. denominator
will be 0 so we are managing vertical lines using a temporary variable.
*/

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        int n=points.size();
        int ans=0;
        for(int i=0;i<n;i++){
            unordered_map<double,int> mp;
            int vertical=0;
            for(int j=i+1;j<n;j++){
                if(points[j][0]-points[i][0] ==0){vertical++;ans=max(ans,vertical);continue;}
                double slope= (double)(points[j][1]-points[i][1])/(points[j][0]-points[i][0]+1.0-1.0);

                mp[slope]++;
                ans=max(ans,mp[slope]);
            }
             mp.clear();

        }
        return ans+1;
    }
};