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Check whether four points make a parallelogram

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  • Difficulty Level : Basic
  • Last Updated : 22 Jun, 2022
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Given four points in a 2-dimensional space we need to find out whether they make a parallelogram or not.  

A parallelogram has four sides. Two opposite sides are parallel and are of same lengths. parallelogram.


Points = [(0, 0),  (4, 0),  (1, 3),  (5, 3)]
Above points make a parallelogram.

Points = [(0, 0), (2, 0), (4, 0), (2, 2)]
Above points does not make a parallelogram 
as first three points itself are linear.

Problems for checking square and rectangle can be read from Square checking and Rectangle checking but in this problem, we need to check for the parallelogram. The main properties of the parallelogram are that opposite sides of parallelogram are parallel and of equal length and diagonals of parallelogram bisect each other. We use the second property to solve this problem. As there are four points, we can get total 6 midpoints by considering each pair. Now for four points to make a parallelogram, 2 of the midpoints should be equal and rest of them should be different. In below code, we have created a map, which stores pairs corresponding to each midpoint. After calculating all midpoints, we have iterated over the map and check the occurrence of each midpoint, If exactly one midpoint occurred twice and other have occurred once, then given four points make a parallelogram otherwise not. 


// C++ code to test whether four points make a
// parallelogram or not
#include <bits/stdc++.h>
using namespace std;
// structure to represent a point
struct point {
    double x, y;
    point() { }
    point(double x, double y)
        : x(x), y(y) { }
    // defining operator < to compare two points
    bool operator<(const point& other) const
        if (x < other.x) {
            return true;
        } else if (x == other.x) {
            if (y < other.y) {
                return true;
        return false;
// Utility method to return mid point of two points
point getMidPoint(point points[], int i, int j)
    return point((points[i].x + points[j].x) / 2.0,
                (points[i].y + points[j].y) / 2.0);
// method returns true if point of points array form
// a parallelogram
bool isParallelogram(point points[])
    map<point, vector<point> > midPointMap;
    // looping over all pairs of point to store their
    // mid points
    int P = 4;
    for (int i = 0; i < P; i++) {
        for (int j = i + 1; j < P; j++) {
            point temp = getMidPoint(points, i, j);
            // storing point pair, corresponding to
            // the mid point
            midPointMap[temp].push_back(point(i, j));
    int two = 0, one = 0;
    // looping over (midpoint, (corresponding pairs))
    // map to check the occurrence of each midpoint
    for (auto x : midPointMap) {
        // updating midpoint count which occurs twice
        if (x.second.size() == 2)
        // updating midpoing count which occurs once
        else if (x.second.size() == 1)
        // if midpoint count is more than 2, then
        // parallelogram is not possible
            return false;    
    // for parallelogram, one mid point should come
    // twice and other mid points should come once
    if (two == 1 && one == 4)
        return true;
    return false;
// Driver code to test above methods
int main()
    point points[4];
    points[0] = point(0, 0);
    points[1] = point(4, 0);
    points[2] = point(1, 3);
    points[3] = point(5, 3);
    if (isParallelogram(points))
        cout << "Given points form a parallelogram";
        cout << "Given points does not form a "
    return 0;


Given points form a parallelogram

Time Complexity: O(p2logp) , where p is number of points
Auxiliary Space: O(p2), where p is number of points

This article is contributed by Aarti_Rathi and Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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