Let $C=\{ax+by:a,b\ge0,a+b=1,x,y\in A\}$
Is enough to show that $C= $ conv$(A)$
How can we prove that C is convex ?
I'm afraid this strategy is a dead end, because $C$ is not necessarily convex! To see this, you'll need a vector space of dimension at least 3. So if you're trying to visualize the problem using real vector spaces, you'll have to go beyond the plane.
Also, make sure you re-read your textbook's definition of the convex hull of $A$. It should be the set of all convex combinations of points from $A$, and this means all combinations of $n$ points, not just combinations of 2 points. Now, given this definition, you can prove that conv($A$) is balanced in a straightforward manner. Let $x$ be a point in conv($A$), represent it in terms of $A$, and use this representation to prove that $\lambda x$ is also in conv($A$).