Solution Breakdown

To form a triangle, we need to select 3 points. The condition is that these 3 points cannot all be on the same straight line (they must be non-collinear). We have two lines with points:

• Line 1 has 10 points.
• Line 2 has 11 points.

A triangle can be formed in two distinct ways:

1. By selecting 2 points from Line 1 and 1 point from Line 2.
2. By selecting 2 points from Line 2 and 1 point from Line 1.

We will use the combination formula, $C(n, k) = \frac{n!}{k!(n-k)!}$, to calculate the number of ways for each case.

Case 1: 2 points from Line 1 (10 points) and 1 point from Line 2 (11 points)

• Ways to choose 2 points from 10: $C(10, 2) = \frac{10 \times 9}{2 \times 1} = 45$
• Ways to choose 1 point from 11: $C(11, 1) = 11$
• Number of triangles for this case: $45 \times 11 = 495$

Case 2: 2 points from Line 2 (11 points) and 1 point from Line 1 (10 points)

• Ways to choose 2 points from 11: $C(11, 2) = \frac{11 \times 10}{2 \times 1} = 55$
• Ways to choose 1 point from 10: $C(10, 1) = 10$
• Number of triangles for this case: $55 \times 10 = 550$

Total Number of Triangles

To find the total number of possible triangles, we add the results from both cases.

Total Triangles = (Triangles from Case 1) + (Triangles from Case 2)

Total Triangles = $495 + 550 = \mathbf{1045}$

This corresponds to option C.