Joint support of a bivariate random vector I encountered a statistics problem that seems basic and I feel that I am being tripped by the graph of part c). Since part c) is the only part I feel I need help with, I will only mention what I got for parts a,b, and c.
Here's the wording of the problem:
Suppose that $(X,Y)$ is a continuous bivariate random vector with support $A=(0,1)\times (0,1)$. Set $U=2X+Y$ and $V=X-2Y$.
a) What is the (marginal) support of the random variable $U$?
b) What is the (marginal) support of the random variable $V$?
c) Sketch a graph showing the (joint) support of the bivariate random vector (U,V). Label the axes as well as any functions that you draw.
I know that due to the support of $X$ and $Y$ being within the interval $(0,1)$ then the support of $U$ and $V$ should be $(0,3)$ and  $(-2,1)$ respectively.
I may be overthinking this and I just want to know what should I change to get the correct joint support (since I am currently not confident in what I have).
As my result after setting X=1 and X=0, and then Y=1 and Y=0 I got that the region with U and V as the horizontal and vertical axis would be bounded by the quadrilateral figure
$V=\frac{1}{2}U$ for $U$ in $(0,2)$
$V=\frac{1}{2}U-\frac{3}{2}$ for $U$ in $(1,3)$
$V=-U$ for $U$ in $(0,1)$
$V=3-U$ for $U$ in $(2,3)$

any feedback is much appreciated.

A simulation experiment shows this is not correct:
X=runif(1e6)
Y=runif(1e6)
plot(2*X+Y,X-2*Y)

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If one inverts the definition of $(U,V)$
$$\left(\matrix{2 &1\\ 1 &-2\\}\right)\left(\matrix{X\\Y\\}\right)=\left(\matrix{U\\V\\}\right)$$
into
$$\left(\matrix{.4 &.2\\ .2 &-.4\\}\right)\left(\matrix{U\\V\\}\right)=\left(\matrix{X\\Y\\}\right)$$
the constraints $$0\le X\le 1\quad0\le Y\le 1$$
turn into
$$0\le 0.4U+0.2V\le 1\quad 0\le0.2U-0.4V\le1$$
Which provides the same quadrilatere
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