We need to follow two rules, the first one is no points are allowed to be the same. The second one is we cannot have any points on the same line. Hence B(1, 1) and C(2, 2) cannot be included.

There are 25 points from 5*5 but we cannot include (0, 0). Thus we want to choose any 2 points from the rest of the 24. \(24C2 = 24*23/2 = 12*23 = 276\).

Next, we eliminate the points on the same line. Any 2 points from (x, x) cannot be chosen so that is \(4C2 = 6\) options ruled out. Don't count (0, 0) as one of the points since we already discluded it. We also have (1, 2) + (2, 4), and (2, 1) + (4, 2) so that is 2 more triangles. Finally, we cannot have the other 2 points both on the x-axis or y-axis so we ruled out \(4C2 * 2 = 12\) triangles. Therefore we take out \(12 + 6 + 2 = 20\) triangles in total.

\(276 - 20 = 256\).

Ans: D
_________________