Why does $\tilde{a} + \vec{v_1} = \tilde{b} = \tilde{a} + \vec{v_2}$ imply that $\vec{v_1} = \vec{v_2}$ in an affine space?

Question

Let $A$ be an affine space.

Question: Why does $\tilde{a} + \vec{v_1} = \tilde{b} = \tilde{a} + \vec{v_2}$ imply that $\vec{v_1} = \vec{v_2}$?

Specifically, why does this follow directly from the definitional fact that, in an affine space $A$, the group action is assumed to be free and transitive?

(I'm pretty sure this follows from the fact that the group action is free means that $\tilde{a} + \vec{v} = \tilde{a} \implies \vec{v} = \vec{0}$, but I don't know how to go from this to the conclusion).

Answer 1

Note that $\tilde b+(\vec{v_2}-\vec{v_1}) = \tilde a+\vec{v_1}+(\vec{v_2}-\vec{v_1})=\tilde a+\vec{v_2}=\tilde b$. So $\vec{v_2}-\vec{v_1}=0$.

Answer 2

This is because the action of the vector space on $A$ is simply transitive.