Let $ \Delta OPQ$ be a triangle under the consideration. Let one of the triangle's $ \Delta OPQ$ vertices be at the origin of the integer coordinates (say $ O(0,0)$ ) , $ C$ be the point inside of $ \Delta OPQ$ and $ (x_1, y_1)$, $ (x_2, y_2)$ and $ (x_3,y_3)$ be the coordinates of $ P$, $ Q$ and $ C$, respectively. Without loss of generality let's say $ 0 \leq x_i \leq y_i$ (it's the same the other way round). Now, we observe that the point $ C$ is the common vertex of three small triangles inside of $ \Delta OPQ$: that is $ \Delta OCP$, $ \Delta CPQ$ and $ \Delta CQO$. Firstly, we notice that from the point $ C$ the points $ O$, $ P$ and $ Q$ are visible, since there are no other lattice points on the sides $ CO$ , $ CP$ and $ CQ$. Secondly, we observe that the points $ P$ and $ Q$ are both visible from the origin $ O$.Then $ gcd (x_i;y_i)=1$, and so $ 0\leq \frac {x_i}{y_i} \leq1$ for $ i= 1,2,3$. Therefore, $ \frac {x_i}{ y_i}$ for any $ i$ ( $ i=1,2,3$) belong to the Farey sequence. Hence, $ \Delta OCP$, $ \Delta CPQ$ and $ \Delta CQO$ are the primitive lattice triangles. The area of any of these triangles is the same, equal to $ \frac {1}{2}$. So, the lattice point $ C$ is the center of gravity.

Using Pick's theorem, the area of $ ABC$ is $ \frac{3}{2}$, The area of each triangle $ \triangle{APC},\triangle{APB},\triangle{BPC}$ is $ \frac{1}{2}$ hence, point $ P$ divides the area in three equal parts, therefore it must be the baricenter.