Line $L_1$ of slope 2 and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin O . In the first…
Line $L_1$ of slope 2 and line $L_2$ of slope $\frac{1}{2}$ intersect at the origin O . In the first quadrant, $\mathrm{P}_1, \mathrm{P}_2, \ldots . \mathrm{P}_{12}$ 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 $\mathrm{O}, \mathrm{P}_1, \mathrm{P}_2, \ldots \mathrm{P}_{12}$, $\mathrm{Q}_1, \mathrm{Q}_2, \ldots . \mathrm{Q}_9$, is:
$1080$
$1134$
$1026$
$1188$
Solution
Total number of $\Delta$ are $\begin{aligned} & ={ }^9 \mathrm{C}_1{ }^{12} \mathrm{C}_2+{ }^9 \mathrm{C}_2{ }^{12} \mathrm{C}_1+{ }^1 \mathrm{C}_1{ }^9 \mathrm{C}_1{ }^{12} \mathrm{C}_1 \\
& =594+432+108 \\
& =1134 \end{aligned}$
.