Find an Integer point on a line segment with given two ends

Given two points pointU and pointV in XY-space, we need to find a point which has integer coordinates and lies on a line going through points pointU and pointV.
Examples:

If  pointU = (1, -1 and pointV = (-4, 1)
then equation of line which goes 
through these two points is,
2X + 5Y = -3
One point with integer co-ordinate which
satisfies above equation is (6, -3)

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can see that once we found the equation of line, this problem can be treated as Extended Euclid algorithm problem, where we know A, B, C in AX + BY = C and we want to find out the value of X and Y from the equation.
In above Extended Euclid equation, C is gcd of A and B, so after finding out the line equation from given two points if C is not a multiple of gcd(A, B) then we can conclude that there is no possible integer coordinate on the specified line. If C is a multiple of g, then we can scale up the founded X and Y coefficients to satisfy the actual equation, which will be our final answer.

//   C++ program to get Integer point on a line 
#include <bits/stdc++.h> 
using namespace std; 
  
//  Utility method for extended Euclidean Algorithm 
int gcdExtended(int a, int b, int *x, int *y) 
{ 
    // Base Case 
    if (a == 0) 
    { 
        *x = 0; 
        *y = 1; 
        return b; 
    } 
  
    int x1, y1; // To store results of recursive call 
    int gcd = gcdExtended(b%a, a, &x1, &y1); 
  
    // Update x and y using results of recursive 
    // call 
    *x = y1 - (b/a) * x1; 
    *y = x1; 
  
    return gcd; 
} 
  
//  method prints integer point on a line with two 
// points U and V. 
void printIntegerPoint(int c[], int pointV[]) 
{ 
    //  Getting coefficient of line 
    int A = (pointU[1] - pointV[1]); 
    int B = (pointV[0] - pointU[0]); 
    int C = (pointU[0] * (pointU[1] - pointV[1]) + 
             pointU[1] * (pointV[0] - pointU[0])); 
  
    int x, y;  // To be assigned a value by gcdExtended() 
    int g = gcdExtended(A, B, &x, &y); 
  
    // if C is not divisible by g, then no solution 
    // is available 
    if (C % g != 0) 
        cout << "No possible integer point\n"; 
  
    else
  
        //  scaling up x and y to satisfy actual answer 
        cout << "Integer Point : " << (x * C/g) << " "
             << (y * C/g) << endl; 
} 
  
//   Driver code to test above methods 
int main() 
{ 
    int pointU[] = {1, -1}; 
    int pointV[] = {-4, 1}; 
  
    printIntegerPoint(pointU, pointV); 
    return 0; 
} 

Output:

Integer Point : 6 -3

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. 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.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Recommended Posts: