Draw some more linear combinations of geometric vectors

Given vectors $A = (2,1), \ B = (1,3)$ , and $C = (xA + yB)$ for scalars $x$ and $y$ :

Plot the geometric vectors from $(0,0)$ to $C$ for the values
$\begin{align*} x &= \frac{1}{2} & y &= \frac{1}{2} \\ x &= \frac{1}{4} & y &= \frac{3}{4} \\ x &= \frac{1}{3} & y &= \frac{2}{3} \\ x &= 2 & y &= -1 \\ x &= 3 & y &= -2 \\ x &= -\frac{1}{2}, & y &= \frac{3}{2} \\ x &= -1 & y &= 2. \end{align*}$
If we let $x$ and $y$ run through all pairs of real numbers such that $x + y = 1$ what is the set of points $C$ ? (Just give a graphical view, no proof necessary.)
If we consider all pairs of points $x$ and $y$ with $x \in [0,1]$ and $y \in [0,1]$ make a sketch of the set of points $C$ we obtain.
What is the set of all points $C$ if $x \in [0,1]$ and $y \in \mathbb{R}$ ?
What is the set of all points $C$ if we allow $x$ and $y$ to be any real numbers?

The plot is as follows (with the locus of points $C$ for part (b) in red):

The points for each value of $t$ are

$\begin{align*} x &= \frac{1}{2} & y &= \frac{1}{2} & \implies && C &= \left( \frac{3}{2}, 2 \right) \\ x &= \frac{1}{4} & y &= \frac{3}{4} & \implies && C &= \left( \frac{5}{4}, \frac{5}{2} \right) \\ x &= \frac{1}{3} & y &= \frac{2}{3} & \implies && C &= \left( \frac{4}{3}, \frac{7}{3} \right) \\ x &= 2 & y &= -1 & \implies && C &= \left( 3, -1 \right) \\ x &= 3 & y &= -2 & \implies && C &= \left( 4, -3 \right) \\ x &= -\frac{1}{2} & y &= \frac{3}{2} & \implies && C &= \left( \frac{1}{2}, 4 \right) \\ x &= -1 & y &= 2 & \implies && C &= \left( 0, 5 \right) \\ \end{align*}$
The set of points $C$ as $x$ and $y$ run through real numbers such that $x+y = 1$ forms a line. It is drawn in red in the plot in part (a).
The set of points in the parallelogram formed by the origin, $A$ , $B$ , and $A+B$ .
The horizontal band obtained by adding $xA$ to the line $y = \frac{1}{3} x$ for each $0 \leq x \leq 1$ .
The whole plane $\mathbb{R}^2$ .