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How many triangles with positive area are there whose vertices are points in the $xy$-plane whose coordinates are integers $(x, y)$ satisfying $1 \le x \le 4$ and $1 \le y \le 4$?


There are $16$ total points in this range. Any triangle is formed by choosing $3$ out of these sixteen points. If that were the case, then the answer would trivially be $\binom{16}{3}=560$.

But alas, degenerate triangles (where all three points are collinear or all points have the same coordinates) have an area of $0$, which forbids them from contributing to the answer.

How can I solve for the number of degenerate triangles?

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The number of straight lines which contain at least 3 points of the collection is 14. (4 horizontal, 4 vertical, 6 diagonal)

The 4 horizontal and 4 vertical lines contain 4 points each. Hence the number of ways to choose 3 collinear points from these 8 lines is $8 \cdot {4 \choose 3}$.

The 2 main diagonals contain 4 points each. Hence the number of ways to choose 3 collinear points from these 2 lines $2 \cdot {4 \choose 3}$.

The other 4 diagonals contain 3 points each. Hence the number of ways to choose 3 collinear points from these 4 lines $4 \cdot {3 \choose 3}$ collinear points.

Since none of these lines have 3 points in common, we have not over-counted. Hence the number of degenerate triangles are $$ 8 \cdot {4 \choose 3} + 2 \cdot {4 \choose 3} + 4 \cdot {3 \choose 3} $$

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