# 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)

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.

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