Pick's Theorem trivializes the problem ( doesn't it?)

Method 1. is by simple application of Pick theorem. $ A(\Delta)= i(\Delta) + \frac{1}{2} b(\Delta) - 1$ so $ A(\Delta)= \frac{3}{2} -1= \frac{1}{2}$ Method 2 is as follows: There is a triangle $ \Delta OPQ$ with $ O(0,0)$ , $ P(x_1,y_1)$ and $ Q(x_2,y_2)$. We know that points $ P(x_1,y_1)$ and $ Q (x_2,y_2)$ are visible from the origin $ O(0,0)$. So, $ (x_1, y_1)=1$ , $ (x_2,y_2)=1$. That means that $ \frac{x_1}{y_1}$, and $ \frac{x_2}{y_2}$ are irreductible. So, $ \frac{x_1}{y_1}$ and $ \frac{x_2}{y_2}$ are the terms of Farey sequence. But if $ \frac{x_1}{y_1}$, and $ \frac{x_2}{y_2}$ are two successive terms of Farey sequence, then $ x_1 \cdot y_2 - x_2 \cdot y_1=1$. From the other part a surface of an arbitrary trangle is given by the formula $ A(\Delta)= \frac{1}{2} \mid x_1 \cdot y_2 - x_2 \cdot y_1 \mid$ so $ A(\Delta OPQ ) = \frac{1}{2}$ and it is done.