A triangle must have positive area. The only case it doesn't, is when the three points are collinear(on the same straight line).

And, your math teacher isn't evil. There are only 10 lattice points for us to choose.

We only need to consider how many combinations of those 3 points are collinear, subtract that from the total number of possible combinations of choosing 3 points out of 10 points, and we get our answer.

Firstly, number of ways we can choose 3 points from 10 points = \(^{10}C_3 = 120\).

Consider any straight line that passes through 3 or more points of the pattern.

There are 2 vertical lines, 2 horizontal lines, and 2 oblique lines that passes through 3 or more points of the pattern.

One of each type of lines passes through 4 points of the pattern, and another of each type of lines passes through 3 points of the pattern.

Let's call those lines which passes through 4 points "Type A" lines, and those which doesn't "Type B" lines.

Number of ways that all 3 points lies on a "Type A" line = \(^4C_3 = 4\)

There are 3 "Type A" lines: \(3\left(^4C_3\right) = 12\)

Number of ways that all 3 points lies on a "Type B" line = \(^3C_3 = 1\)

There are 3 "Type B" lines: \(3\left(^3C_3\right) = 3\)

Now, subtract these cases from the total number of cases.

Number of ways that 3 points in the pattern are chosen such that they form a triangle with positive area

= 120 - 12 - 3

= **105**

In case you don't know these, you can list the possible ways and count them one by one. There are only 105 cases so it should be fast enough.