Line L1 of slope 2 and line L2 of slope 12 intersect at the origin O. In the first quadrant, P1, P2, , P12 are 12 points on line L1 and Q1, Q2, , Q9 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, P1, P2, , P12, Q1, Q2, , Q9, is:
- 1026
- 1188
- 1134
- 1080
Answer
(3) 1134
Solution
Collinear sets: O with P1, , P12: 13 points on L1;
O with Q1, , Q9: 10 points on L2.
Number of triangles = (all selections of 3 points) − (selections of 3 collinear points)
= 22C3 - 13C3 - 10C3
= 1540 - 286 - 120
= 1134
Answer: (3)