Points $A (b, 2c), B(4 b, 3c), C (5b, c)$, and $D(2b, 0)$ form a quadrilateral. How would you classify the Quadrilateral and explain your steps? I have homework -

Points $A (b, 2c), B (4 b, 3c), C (5b, c)$, and $D (2b, 0)$ form a
quadrilateral. How would you classify the Quadrilateral and explain
your steps?

But I don't know really how to start?
I mean should I draw figure? And where are points (in a coordinate system)?
If someone can explain?
Thanks again!!!
 A: Let $A':=(1,2),\,B':=(4,3),\,C':=(5,1),\,D':=(2,0)$ and we apply a linear transformation $(x,y)\to (bx,cy)$ to them, obtaining $A,\,B,\,C,\,D$.
As it can be clearly seen from the picture

$A'B'C'D'$ is a parallelogram. More rigorously, because
$\overrightarrow{A'B'}=\overrightarrow{D'C'}=(3,1)$. It will remain parallelogram after the transform (e.g. for the same reasoning, $\overrightarrow{AB}=\overrightarrow{DC}=(3b,c)$).
A: You have six lines: $AB,BC,CD,DA,AC,BD$.  Four of them are sides and two of them are diagonals.    You can probably determine which are which by drawing a figure and I suspect that will be enough for you class.  However we can do it without a figure by calculating that only the two diagonals intersect.
Figure out the lengths and slopes of the segments.  That will tell you how to clasify them.
$AB$ has slope $\frac {3c-2c}{4b-b}=\frac c{3b}$ and length $\sqrt{(3c-2c)^2 + (4b-b)^2} = \sqrt{c^2 + 9b^2}$.
$BC$ has slope $\frac {c-3c}{5b-4b}=-\frac {2c}b$ and length $\sqrt{(-2c)^2 + (b)^2}=\sqrt {4c^2 + b^2}$.
$CD$ has slope $\frac {0-c}{2b-5b}=\frac{-c}{-3b}=\frac c{3b}$.  This slope is equal to $AB$ so $AB$ is parallel to $CD$. So....  Neither $AB$ nor $CD$ can be diagonals so they must be sides (diagonals intersects all sides and the other diagonal so they can not be parallel to any other of the lines).  So the quadrilateral has two parallel sides.  The length of $CD$ is $\sqrt{(0-c)^2 +(2b-3b)^2} = \sqrt{c^2 + 9b^2}$. This is the same length as $AB$ so the quadrilateral has two parallel sides of equal length.  So the quadrilateral is a parallelogram.
And we can predict the other two sides will be parallel and of equal length.  But lets make sure.
The slope of $DA$ is $\frac {2c-0}{2b-b}=\frac {2c}b$  The is equal to the slope of $BC$ so indeed $DA$ is parallel to $BC$ and the quadrilateral has two pairs of parallel sides.  And as predicted the length of $DA = \sqrt{(2c)^2 + b^2} = \sqrt {4c^2 + b^2} = BC$ so the sides are equal.
So the quadrilateral is a parallelogram.
But is it a rhombus?  Is it a rectangle? is it a square?
That's impossible to tell as we don't know the values of $b$ or $c$.
It would be a rhombus if $AB=BC=CD=DA$ or if $\sqrt{c^2+9b^2}=\sqrt{4c^2 +b^2}$.  That would mean $c^2 +9b^2 = 4c^2 + b^2$ or $8b^2 = 3c^2$ or if $b =\sqrt{\frac 38}c$.
And it would be a rectangle of the slopes of $AB$ and $CD$ would be the negative recipricals of the slopes of $BC$ and $DA$.  That is if $\frac c{3b} = -\frac 1{-\frac {2c}b}= \frac b{2c}$.  That would mean $2c^2 = 3b^2$ and that $b=\sqrt{\frac 23}c$.
To be a square both would have to be true and that would be $b =\sqrt{\frac 38}c=\sqrt{\frac 23}c$.  That can only happen if $c=0$ and $b=0$. Which... isn't really fair...(It means all four points are actually one point and... well, the author cheated.)
So ....
It's a parallelogram.  It is a rhombus if $b=\sqrt{\frac 38}c$.  It is a rectangle if $b =\sqrt{\frac 23}c$ and it is not a square.
