Question

Line L1 of slope 2 and line L2 of slope \( \frac{1}{2} \) intersect at the origin O. In the first quadrant, \( P_1, P_2, ..., P_{12} \) are 12 points on line L1 and \( Q_1, Q_2, ..., Q_9 \) are 9 points on line L2. Then the total number of triangles that can be formed having vertices at three of the 22 points O, \( P_1, P_2, ..., P_{12} \), \( Q_1, Q_2, ..., Q_9 \), is:

• A. 1080
• B. 1134
• C. 1026
• D. 1188

Hint:

To count the number of triangles formed by a set of points, consider the cases where the three vertices are chosen such that they are not all on the same line. Here, the lines are L1 and L2, both passing through the origin O. Consider combinations of points taken from these lines.

Answer 1

The Correct Option is B

To form a triangle, we need to choose 3 non-collinear points. The given set of points consists of:

• The origin O (1 point)
• 12 points on line L1 (\( P_1, \dots, P_{12} \))
• 9 points on line L2 (\( Q_1, \dots, Q_9 \))

Total number of points = \( 1 + 12 + 9 = 22 \). We need to consider combinations of 3 points such that they are not collinear. The collinear sets are (O and any two points on L1) and (O and any two points on L2). Case 1: One vertex from L2 and two vertices from L1. Number of ways = \( \binom{9}{1} \times \binom{12}{2} = 9 \times \frac{12 \times 11}{2} = 9 \times 66 = 594 \) Case 2: Two vertices from L2 and one vertex from L1. Number of ways = \( \binom{9}{2} \times \binom{12}{1} = \frac{9 \times 8}{2} \times 12 = 36 \times 12 = 432 \) Case 3: One vertex from L2, one vertex from L1, and the origin O. Number of ways = \( \binom{9}{1} \times \binom{12}{1} \times \binom{1}{1} = 9 \times 12 \times 1 = 108 \) The total number of triangles is the sum of the number of ways in these three cases: Total triangles = \( 594 + 432 + 108 = 1134 \)