JEE Main 2025 (Online) 3rd April Evening ShiftQuestion 1
Line $L_1$ of slope $2$ and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin $O$. In the first quadrant, $P_1,P_2,\ldots,P_{12}$ are $12$ points on line $L_1$ and $Q_1,Q_2,\ldots,Q_9$ 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 $O,P_1,P_2,\ldots,P_{12},Q_1,Q_2,\ldots,Q_9$, is:
Permutation and Combination
Solution
Total triples from 22 points: $\binom{22}{3}=1540$. Collinear triples: $O$ with the 12 points of $L_1$ gives 13 collinear points, so $\binom{13}{3}=286$. Likewise $O$ with the 9 points of $L_2$ gives $\binom{10}{3}=120$. Triangles $=1540-286-120=1134$. Option (C).