Proof on affine combination I have $X:=\{ (x,y) \in [0,1]^2| x = 0 \vee y = 0 \vee x+y=1\} $
I need to show that the set of all affine combinations of $X$ is equal to $\mathbb{R}^2$
Attempt:
by definion of affine combination $x =\sum_i x_i \gamma_i$ and $\sum_i \gamma_i=1$
Therefore I defined any $y \in X : (y_1, y_2) = \gamma_1(x_1,0)+\gamma_2(0,y_2)+(1-\gamma_1-\gamma_2)(x_2,1-x_2)$
But I am missing both the intuition and the last step of the proof. Any idea?
 A: The set of all affine combinations of $X$ is $\operatorname{aff}(X)$ the affine hull of $X$. 
The affine hull of a set $X$ is the smallest affine subset of $\mathbb{R}^2$ which contains $X$.
The affine subsets of $\mathbb{R}^2$ are either a one point set or a line or whole $\mathbb{R}^2$.
If $X$ contained just one point, its affine hull would be $X$ itself, since it is already an affine subspace of $\mathbb{R}^2$.
If $X$ were a subset of a line $L$, which is an affine subspace of $\mathbb{R}^2$, it would hold $\operatorname{aff}(X)=L$.
But here we have a set $X$ which contains three affine linear independent vectors (three vectors which don't lie on a line). And this is enough to get $\operatorname{aff}(X)=\mathbb{R}^2$. You could choose the corner points of your triangle $X$, which are
$$
u_1=\begin{pmatrix}1\\0\end{pmatrix},~u_2=\begin{pmatrix}0\\1\end{pmatrix},~u_3=\begin{pmatrix}0\\0\end{pmatrix}.
$$
A affine combination of $u_1$, $u_2$ and $u_3$ is $c_1u_1+c_2u_2+c_3u_3$ where $c_1+c_2+c_3=1$. To proof that $\operatorname{aff}(X)=\mathbb{R}^2$ we have to show that each point $(x_1,x_2)\in\mathbb{R}^2$ is an affine combination of $u_1$, $u_2$ and $u_3$. We see
$$
\begin{pmatrix}x_1\\x_2\end{pmatrix}=c_1u_1+c_2u_2+c_3u_3=c_1\begin{pmatrix}1\\0\end{pmatrix}+c_2\begin{pmatrix}0\\1\end{pmatrix}+c_3\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}c_1\\c_2\end{pmatrix}.
$$
You see that you can choose $c_1=x_1$ and $c_2=x_2$ and finally $c_3=1-c_1-c_2=1-x_1-x_2$ to get an affine combination.
You may think it looks like a cheat because I chose $u_3=\begin{pmatrix}0\\0\end{pmatrix}$ which made the affine combination independent of $c_3$ such that I could define $c_1$ and $c_2$ totally free. But it is possible to simplify our proof. I could have chosen $u_3=\begin{pmatrix}0\\\frac12\end{pmatrix}$ or each other point which doesn't lie on the line between $u_1$ and $u_2$. If you feel better to use each edge, you can chose
$$
v_1=\begin{pmatrix}\frac12\\0\end{pmatrix},~v_2=\begin{pmatrix}0\\\frac12\end{pmatrix},~v_3=\begin{pmatrix}\frac12\\\frac12\end{pmatrix}.
$$
Using $c_3=1-c_1-c_2$ you get
$$
\begin{pmatrix}x_1\\x_2\end{pmatrix}=c_1\begin{pmatrix}\frac12\\0\end{pmatrix}+c_2\begin{pmatrix}0\\\frac12\end{pmatrix}+c_3\begin{pmatrix}\frac12\\\frac12\end{pmatrix}=\frac12\begin{pmatrix}1-c_2\\1-c_1\end{pmatrix}
$$
and your coefficients are $c_1=1-2x_2$, $c_2=1-2x_1$ and $c_3=2x_1+2x_2-1$.
