imc

We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$, so $\left(\genfrac{}{}{0ex}{}{25}{3}\right)$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2\cdot 1}$, which simplifies to $2300$. Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,5)$ would be on the same line, so $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ is $10$, and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$. We can also count the ones with $4$ on a diagonal. That is $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$, which is 4, and there are $4$ of those diagonals, so that results in $16$. We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $4$. We can also count the ones with a slope of $\frac{1}{2}$, $2$, $-\frac{1}{2}$, or $-2$, with $3$ points in each. Note that there are $3$ such lines, for each slope, present in the grid. In total, this results in $12$. Finally, we subtract all the ones in a line from $2300$, so we have $2300-120-16-4-12=\boxed{2148}$