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 ascertain the quantity of triangles formable from the given points, we must consider the origin \( O \), twelve points \( P_1, P_2, \ldots, P_{12} \) situated on line \( L_1 \), and nine points \( Q_1, Q_2, \ldots, Q_9 \) located on line \( L_2 \).

Step 1: Point Configuration Analysis

• Line \( L_1 \), with equation \( y = 2x \), hosts points \( P_1 \) through \( P_{12} \).
• Line \( L_2 \), with equation \( y = \frac{1}{2}x \), hosts points \( Q_1 \) through \( Q_9 \).
• Both lines intersect at the origin \( O (0, 0) \).

Step 2: Initial Triangle Count Calculation

• The aggregate number of points available is \( 1 \) (origin) \( + 12 \) (on \( L_1 \)) \( + 9 \) (on \( L_2 \)) \( = 22 \).
• The total combinations for selecting 3 points from these 22 to potentially form a triangle are calculated as:
• \(\binom{22}{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540\)

Step 3: Subtraction of Collinear Point Combinations

• The 12 points on line \( L_1 \) are collinear. Combinations of 3 points from these 12 are:
• \(\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\)
• The 9 points on line \( L_2 \) are collinear. Combinations of 3 points from these 9 are:
• \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)
• Combinations involving the origin and two points from the same line are also collinear. These include:
• 2 points from \( L_1 \) and the origin \( O \): \(\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66\)
• 2 points from \( L_2 \) and the origin \( O \): \(\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36\)

Total Collinear Cases Exclusion:

The sum of all non-triangle-forming collinear point combinations is \( 220 + 84 + 66 + 36 = 406 \).

The final count of valid triangles is derived by subtracting these collinear cases from the total combinations:

• Valid Triangles = Total Combinations - Collinear Combinations
• \(1540 - 406 = 1134\)

Conclusion: A total of 1134 triangles can be formed.