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Check whether a straight line can be formed using N co-ordinate points

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Given an array arr[] of N co-ordinate points, the task is to check whether a straight line can be formed using these co-ordinate points.

Input: arr[] = {{0, 0}, {1, 1}, {2, 2}} 
Output: Yes 
Explanation: 
Slope of every two points is same. That is 1. 
Therefore, a straight line can be formed using these points.

Input: arr[] = {{0, 1}, {2, 0}} 
Output: Yes 
Explanation: 
Two points in co-ordinate system always forms a straight line. 
 

Approach: The idea is to find the slope of line between every pair of points in the array and if the slope of every pair of point is same, then these points together forms a straight line. 

// Slope of line formed by 
// two points (y2, y1), (x2, x1)

Slope of Line = y2 - y1
               ---------
                x2 - x1

Below is the implementation of the above approach: 

C++




// C++ implementation to check
// if a straight line
// can be formed using N points
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to check if a straight line
// can be formed using N points
bool isStraightLinePossible(
    vector<pair<int, int> > arr, int n)
{
    // First pair of point (x0, y0)
    int x0 = arr[0].first;
    int y0 = arr[0].second;
 
    // Second pair of point (x1, y1)
    int x1 = arr[1].first;
    int y1 = arr[1].second;
 
    int dx = x1 - x0, dy = y1 - y0;
     
    // Loop to iterate over the points
    for (int i = 0; i < n; i++) {
        int x = arr[i].first, y = arr[i].second;
        if (dx * (y - y1) != dy * (x - x1)){
            cout << "NO";
            return false;
        }
    }
    cout << "YES";
    return true;
}
 
// Driver Code
int main()
{
    // Array of points
    vector<pair<int, int> > arr =
    { { 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 } };
    int n = 4;
     
    // Function Call
    isStraightLinePossible(arr, n);
    return 0;
}


Java




// Java implementation to check
// if a straight line can be
// formed using N points
import java.util.*;
 
class GFG{
 
static class pair
{
    int first, second;
    pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossible(
       ArrayList<pair> arr, int n)
{
     
    // First pair of point (x0, y0)
    int x0 = arr.get(0).first;
    int y0 = arr.get(0).second;
 
    // Second pair of point (x1, y1)
    int x1 = arr.get(1).first;
    int y1 = arr.get(1).second;
 
    int dx = x1 - x0, dy = y1 - y0;
 
    // Loop to iterate over the points
    for(int i = 0; i < n; i++)
    {
        int x = arr.get(i).first;
        int y = arr.get(i).second;
        if (dx * (y - y1) != dy * (x - x1))
        {
            System.out.println("NO");
        }
    }
    System.out.println("YES");
}
 
// Driver code
public static void main(String[] args)
{
     
    // Array of points
    ArrayList<pair> arr = new ArrayList<>();
    arr.add(new pair(0, 0));
    arr.add(new pair(1, 1));
    arr.add(new pair(3, 3));
    arr.add(new pair(2, 2));
 
    int n = 4;
 
    // Function Call
    isStraightLinePossible(arr, n);
}
}
 
// This code is contributed by offbeat


Python3




# Python3 implementation to check
# if a straight line can be formed
# using N points
 
# Function to check if a straight line
# can be formed using N points
def isStraightLinePossible(arr, n):
     
    # First pair of point (x0, y0)
    x0 = arr[0][0]
    y0 = arr[0][1]
 
    # Second pair of point (x1, y1)
    x1 = arr[1][0]
    y1 = arr[1][1]
 
    dx = x1 - x0
    dy = y1 - y0
     
    # Loop to iterate over the points
    for i in range(n):
        x = arr[i][0]
        y = arr[i][1]
         
        if (dx * (y - y1) != dy * (x - x1)):
            print("NO", end = "")
            return False
 
    print("YES", end = "")
    return True
 
# Driver code
 
# Array of points
arr = [ [ 0, 0 ], [ 1, 1 ],
        [ 3, 3 ], [ 2, 2 ] ]
n = 4
 
# Function Call
isStraightLinePossible(arr, n)
 
# This code is contributed by divyeshrabadiya07


C#




// C# implementation to check
// if a straight line can be
// formed using N points
using System;
using System.Collections.Generic;
 
class GFG{
 
public class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossible(
    List<pair> arr, int n)
{
     
    // First pair of point (x0, y0)
    int x0 = arr[0].first;
    int y0 = arr[0].second;
 
    // Second pair of point (x1, y1)
    int x1 = arr[1].first;
    int y1 = arr[1].second;
 
    int dx = x1 - x0, dy = y1 - y0;
 
    // Loop to iterate over the points
    for(int i = 0; i < n; i++)
    {
        int x = arr[i].first;
        int y = arr[i].second;
         
        if (dx * (y - y1) != dy * (x - x1))
        {
            Console.WriteLine("NO");
        }
    }
    Console.WriteLine("YES");
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Array of points
    List<pair> arr = new List<pair>();
     
    arr.Add(new pair(0, 0));
    arr.Add(new pair(1, 1));
    arr.Add(new pair(3, 3));
    arr.Add(new pair(2, 2));
 
    int n = 4;
 
    // Function call
    isStraightLinePossible(arr, n);
}
}
 
// This code is contributed by Princi Singh


Javascript




<script>
// Javascript implementation of the above approach
// Function to check if a straight line
// can be formed using N points
function isStraightLinePossible( arr, n)
{
    // First pair of point (x0, y0)
    var x0 = arr[0][0];
    var y0 = arr[0][1];
 
    // Second pair of point (x1, y1)
    var x1 = arr[1][0];
    var y1 = arr[1][1];
 
    var dx = x1 - x0;
    var dy = y1 - y0;
     
    // Loop to iterate over the points
    for (var i = 0; i < n; i++) {
        var x = arr[i][0], y = arr[i][1];
        if (dx * (y - y1) != dy * (x - x1)){
            document.write("NO");
            return false;
        }
    }
    document.write("YES");
    return true;
}
 
// Driver Code
// Array of points
var arr = [[ 0, 0 ], [ 1, 1 ], [ 3, 3 ], [ 2, 2 ]];
var n = 4;
 
// Function Call
isStraightLinePossible(arr, n);
</script>


Output: 

YES

 

Time Complexity: O(N)

Auxiliary Space: O(1)
 



Last Updated : 07 Jan, 2024
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