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 solve this problem, we need to determine the number of triangles that can be formed with vertices among the points: the origin \( O \), the 12 points \( P_1, P_2, \ldots, P_{12} \) on line \( L_1 \), and the 9 points \( Q_1, Q_2, \ldots, Q_9 \) on line \( L_2 \).

Step 1: Understanding the positions of the points

• Line \( L_1 \) has a slope of 2, which means its equation is \( y = 2x \). Points \( P_1, P_2, \ldots, P_{12} \) lie on this line.
• Line \( L_2 \) has a slope of \( \frac{1}{2} \), which means its equation is \( y = \frac{1}{2}x \). Points \( Q_1, Q_2, \ldots, Q_9 \) lie on this line.
• Both lines intersect at the origin \( O (0, 0) \).

Step 2: Calculating the possible triangles

• The total number of points is \( 1 + 12 + 9 = 22 \) (including the origin).
• The total number of ways to choose any 3 points from these 22 to form a triangle is computed using combinations:
• \(\binom{22}{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540\)

Step 3: Excluding collinear points

• Points \( (P_1, P_2, \ldots, P_{12}) \) on line \( L_1 \) are collinear. The number of ways to choose 3 points from these 12 is:
• \(\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\)
• Points \( (Q_1, Q_2, \ldots, Q_9) \) on line \( L_2 \) are collinear. The number of ways to choose 3 points from these 9 is:
• \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)
• Points \( (O, P_i, P_j) \) or \( (O, Q_k, Q_l) \) where the points are on the same line are also collinear. The number of such cases (2 points on the same line plus \( O \)) is:
• \(\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66\)for line \( L_1 \)
• \(\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36\)for line \( L_2 \)

Excluding all collinear cases:

Total collinear cases = \( 220 + 84 + 66 + 36 = 406 \)

Thus, the number of triangles that can be formed is:

• Total triangles = Total combinations - Collinear cases
• \(1540 - 406 = 1134\)

Conclusion: The total number of triangles that can be formed is 1134.

Answer 2

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 = \( 594 + 432 + 108 = 1134 \)