Line L_1 of slope 2 and line L_2 of slope 12 intersect at the origin O.

Question

Line $L_1$ of slope 2 and line $L_2$ of slope $\dfrac12$ intersect at the origin O. In the first quadrant, $P_1,P_2,\ldots,P_{12}$ are 12 points on line $L_1$ and $Q_1,\ldots,Q_9$ are 9 points on line $L_2$. Then the total number of triangles that can be formed having vertices at three of the 22 points $O,P_1,\ldots,P_{12},Q_1,\ldots,Q_9$ is:

Accepted Answer

Points on $L_1$ (with O) total 13 collinear points; points on $L_2$ (with O) total 10 collinear points. A triangle needs one point from each of two different lines (plus possibly O), or points not all collinear.

Valid triangles: (2 points from $P$'s, 1 point from $Q$'s) + (1 point from $P$'s, 2 points from $Q$'s) + (1 point from $P$'s, 1 from $Q$'s, and O):

$$\binom{12}{2}\binom{9}{1} + \binom{12}{1}\binom{9}{2} + \binom{12}{1}\binom{9}{1}\binom{1}{1} = 594+432+108 = 1134$$

Solutions · 1