Line $$L_1$$ of slope 2 and line $$L_2$$ of slope $$\dfrac{1}{2}$$ 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:

We have a total of 22 points:
  • the origin $$O$$,
  • $$12$$ points $$P_1, P_2,\,\ldots,\,P_{12}$$ on the line $$L_1$$ (slope $$2$$),
  • $$9$$ points $$Q_1, Q_2,\,\ldots,\,Q_{9}$$ on the line $$L_2$$ (slope $$\tfrac12$$).

A triangle can be formed by any three non-collinear points. Hence

Total number of triangles
$$=\;\bigl(\text{all possible triples of points}\bigr)\;-\;\bigl(\text{collinear triples}\bigr).$$

Step 1: Count all possible triples of points
There are $$22$$ points in all, so the number of ways to choose any three of them is
$$\binom{22}{3} \;=\; \frac{22 \times 21 \times 20}{6} \;=\; 1540.$$

Step 2: Subtract triples that are collinear

• Triples on line $$L_1$$: Along $$L_1$$ we have the origin $$O$$ plus the 12 points $$P_i$$, altogether $$13$$ collinear points. The number of triples drawn entirely from these $$13$$ points is
$$\binom{13}{3} \;=\; \frac{13 \times 12 \times 11}{6} \;=\; 286.$$

• Triples on line $$L_2$$: Along $$L_2$$ we have the origin $$O$$ plus the 9 points $$Q_j$$, altogether $$10$$ collinear points. The number of triples drawn entirely from these $$10$$ points is
$$\binom{10}{3} \;=\; \frac{10 \times 9 \times 8}{6} \;=\; 120.$$

Step 3: Adjust for any overlap
A triple can lie on both $$L_1$$ and $$L_2$$ only if all three of its points are common to the two lines. The only common point of $$L_1$$ and $$L_2$$ is the origin $$O$$, so no triple is counted in both collinear groups. Therefore, no further adjustment is required.

Step 4: Final count of triangles
$$\begin{aligned} \text{Number of triangles} &= 1540 \;-\; 286 \;-\; 120 \\ &= 1134. \end{aligned}$$

Hence, the total number of distinct triangles that can be formed is $$1134$$, which corresponds to Option B.