Find the equation of the straight line passing through the given points

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Given an array arr containing N coordinate points in a plane, the task is to check whether the coordinate points lie on a straight line or not. If they lie on a straight line then print Yes and also the equation of that line otherwise print No.

Example:

Input: arr[] = {{1, 1}, {2, 2}, {3, 3}}
Output:
Yes
1x- 1y=0

Input: arr[] = {{0, 1}, {2, 0}}
Output:
Yes
2y+x-2 = 0

Input: arr[] = {{1, 5}, {2, 2}, {4, 6}, {3, 5}}
Output: No

Approach: The idea is to find the equation of the line that can be formed using any one pair of points given in the array and if all other points satisfy the equation of the line formed using the pair of the points, then all these points together form a straight line. So, if all points satisfy the equation of the line, then print Yes followed by the equation of the line, otherwise print No.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a straight line
// can be formed using N points
void isStraightLinePossibleEq(
    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,
        c = dy * x0 - dx * y0;
 
    // Loop to iterate over the points
    // and check whether they satisfy
    // the equation or not.
    for (int i = 2; i < n; i++) {
        int x = arr[i].first, y = arr[i].second;
        if ((dx * y) - (dy * x) != c) {
            cout << "No";
            return;
        }
    }
    cout << "Yes" << endl;
    cout << dy << "x-" << dx
         << "y=" << c << "\n";
}
 
// Driver Code
int main()
{
    // Array of points
    vector<pair<int, int> > arr
        = { { 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 } };
    int N = 2;
 
    // Function Call
    isStraightLinePossibleEq(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
public class GFG
{
 
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossibleEq(int arr[][], int n)
{
    // First pair of point (x0, y0)
    int x0 = arr[0][0];
    int y0 = arr[0][1];
 
    // Second pair of point (x1, y1)
    int x1 = arr[1][0];
    int y1 = arr[1][1];
 
    int dx = x1 - x0,
        dy = y1 - y0,
        c = dy * x0 - dx * y0;
 
    // Loop to iterate over the points
    // and check whether they satisfy
    // the equation or not.
    for (int i = 2; i < n; i++) {
        int x = arr[i][0], y = arr[i][1];
        if ((dx * y) - (dy * x) != c) {
            System.out.print("No");
            return;
        }
    }
    System.out.print("Yes" + "\n");
    System.out.print(dy + "x-" + dx
         + "y=" + c + "\n");
}
 
// Driver Code
public static void main(String args[])
{
     
    // Array of points
    int arr[][] = {{ 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 }};
    int N = 2;
 
    // Function Call
    isStraightLinePossibleEq(arr, N);
 
}
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to check if a straight line
# can be formed using N points
def isStraightLinePossibleEq(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
    c = dy * x0 - dx * y0
 
       # Loop to iterate over the points
       # and check whether they satisfy
       # the equation or not.
    for i in range(2, n):
        x = arr[i][0], y = arr[i][1]
        if (dx * y) - (dy * x) != c:
            print("No")
            return
 
    print("Yes")
    print(str(dy)+ "x" +"-" + str(dx)+ "y"+ "="+ str(c))
 
    # Driver Code
 
 
    # Array of points
arr = [[0, 0], [1, 1], [3, 3], [2, 2]]
N = 2
 
# Function Call
isStraightLinePossibleEq(arr, N)
 
# This code is contributed by Potta Lokesh

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossibleEq(int [,]arr, int n)
{
    // First pair of point (x0, y0)
    int x0 = arr[0, 0];
    int y0 = arr[0, 1];
 
    // Second pair of point (x1, y1)
    int x1 = arr[1, 0];
    int y1 = arr[1, 1];
 
    int dx = x1 - x0,
        dy = y1 - y0,
        c = dy * x0 - dx * y0;
 
    // Loop to iterate over the points
    // and check whether they satisfy
    // the equation or not.
    for (int i = 2; i < n; i++) {
        int x = arr[i, 0], y = arr[i, 1];
        if ((dx * y) - (dy * x) != c) {
            Console.Write("No");
            return;
        }
    }
    Console.Write("Yes" + "\n");
    Console.Write(dy + "x-" + dx
         + "y=" + c + "\n");
}
 
// Driver Code
public static void Main()
{
     
    // Array of points
    int[,] arr = new int[4, 2] {{ 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 }};
    int N = 2;
 
    // Function Call
    isStraightLinePossibleEq(arr, N);
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript program for the above approach
 
    // Function to check if a straight line
    // can be formed using N points
    const isStraightLinePossibleEq = (arr, n) => {
     
        // First pair of point (x0, y0)
        let x0 = arr[0][0];
        let y0 = arr[0][1];
 
        // Second pair of point (x1, y1)
        let x1 = arr[1][0];
        let y1 = arr[1][1];
 
        let dx = x1 - x0,
            dy = y1 - y0,
            c = dy * x0 - dx * y0;
 
        // Loop to iterate over the points
        // and check whether they satisfy
        // the equation or not.
        for (let i = 2; i < n; i++) {
            let x = arr[i][0], y = arr[i][1];
            if ((dx * y) - (dy * x) != c) {
                cout << "No";
                return;
            }
        }
        document.write("Yes<br/>");
        document.write(`${dy}x-${dx}y=${c}<br/>`);
    }
 
    // Driver Code
 
    // Array of points
    let arr = [[0, 0], [1, 1], [3, 3], [2, 2]];
    let N = 2;
 
    // Function Call
    isStraightLinePossibleEq(arr, N);
 
    // This code is contributed by rakeshsahni
 
</script>

Output

Yes
1x-1y=0

Time Complexity: O(N)
Auxiliary Space: O(1)

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