I am studying triangles on the plane $\mathbb R^2$ whose vertices have integer coordinates. If any such solid triangle (i.e., the convex hull of the vertices) has no other points with integer coordinates, we call it a primitive triangle.
I want to prove that every primitive triangle has area $1/2$. We may assume without loss of generality that one of the vertices is the origin, and the other two vertices are in the first quadrant $\mathbb N^2$. Let $(a,b)$ and $(x,y)$ be the other two vertices. Clearly, the triangle has area
$$A = \frac 12 \Vert (a,b) \times (x,y) \Vert = \frac 12 |ay - bx|$$
Thus, it suffices to show that $x,y$ are coprime and $a,b$ are coefficients arising from Bézout's identity
$$ax - by = \gcd(x,y) = 1$$
However, I have no idea how to continue from this point. Could someone give me a hint?
Here is a nice algebraic proof. The points with integer coordinates form the additive group $\mathbb Z^2$. A triangle with integer coordinates can be seen as two vectors at the origin. The triangle contains no no other points with integer coordinates iff the vectors generate $\mathbb Z^2$. And this happens iff the matrix with their coordinates has determinant $\pm 1$. So, the area of the triangle is $1/2$.