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.