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.

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)

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 divisble by g, then no solution
    // is available
    if (C % g != 0)
        cout << "No possible integer point ";
        //  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;


Integer Point : 6 -3

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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