The problem can be formulated and solved as a linear program.
Enumeration of the extreme points $\mathbf{x}_1$, ..., $\mathbf{x}_n$ and rays $\mathbf{d}_1$, ..., $\mathbf{d}_m$ yields a $\mathcal{V}$-representation (vertex-representation) of the polyhedron (feasible region) $X$. Just write down two matrices $V$ and $R$, where the rows of $V$ are the vertices $\mathbf{x}_1$, ..., $\mathbf{x}_n$ and the rows of $R$ are the rays $\mathbf{d}_1$, ..., $\mathbf{d}_m$. That is $X=\{\mathbf{x}\mid V\mathbf{u} + R\mathbf{v} = \mathbf{x}, \mathbf{u} \geq \mathbf{0}, \mathbf{1}^T\mathbf{u} = 1, \mathbf{v} \geq \mathbf{0}\}$.
Now, you ask whether some given $\mathbf{x}^*$ is contained in $X$, i.e, find a feasible solution $(\mathbf{u}_0, \mathbf{v}_0)$ of
$\quad$ $V\mathbf{u} + R\mathbf{v} = \mathbf{x}^*, \mathbf{u} \geq \mathbf{0}, \mathbf{1}^T\mathbf{u} = 1, \mathbf{v} \geq \mathbf{0}$.
This can easily be done by linear programming.
Note that the vector $\mathbf{u}_0$ encodes the coefficients of the convex combination of the vertices and the vector $\mathbf{v}_0$ encodes the coefficients of the conical combination of the rays.
Define the Euclidean metric on $S$. Note that for every point of $S$ like $(x_1,x_2)$ such that $x_1+2x_2\le 4$ there exist two points $(x_1+2,x_2-1)$ and $(x_1-2,x_2+1)$ both belonging to $S$ (since $[x_1+2]+2[x_2-1]=x_1+2x_2\le 4$ and $[x_1-2]+2[x_2+1]=x_1+2x_2\le 4$). Since $S$ is convex then whole the line connecting these 2 points belongs to $S$. Now note that $$(x_1,x_2)=\dfrac{1}{2}(x_1-2,x_2+1)+\dfrac{1}{2}(x_1+2,x_2-1)$$which implies that $(x_1,x_2)$ is in the middle of that line and therefore is not an extreme point. Also refer to the following illusion for feasible directions:
Best Answer
In standard format, we get the system: \begin{equation*} \begin{alignedat}{2} -x_1 & {}+{} & x_2&{}+{}&x_3& &&= 2\\ -x_1&{}+{}&2x_2& &&{}+{}&x_4\ &=6 \end{alignedat} \end{equation*} with all variables positive. The matrix of the system is: $$A=\begin{pmatrix}-1&1&1&0\\ -1&2&0&1\end{pmatrix}.$$ Any extreme direction $d$ can be obtained as: $$d=\begin{pmatrix}-B^{-1}a_j\\e_j\end{pmatrix},$$ where $B$ is a $2\times 2$ invertible submatrix of $A$, $a_j$ is the $j$th column of $A$, not in $B$, such that $B^{-1}a_j\leq0$ and $e_j$ is the canonical vector with a one in the position of the column $a_j$.
For example, let $B=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, invertible submatrix of $A$. We have that $B^{-1}a_1=\begin{pmatrix}-1\\-1\end{pmatrix}\leq0$. The canonical vector $e_1$ has a one in the position of $a_1$, i. e., $e_1=\begin{pmatrix}1\\0\end{pmatrix}$. Therefore, we get the direction $$d=\begin{pmatrix}1\\0\\1\\1\\\end{pmatrix},\ \text{writing }-B^{-1}a_1\text{ in the position of }B\text{ and }e_1\text{ in the left entries.}$$ In our original system, by erasing the variables $x_3$ and $x_4$, we get the extreme direction $\begin{pmatrix}1\\0\end{pmatrix}$.