How do you show whether the given is a Vector Space over $\mathbb R$? I'm stuck on a HW problem for my Linear Algebra class. 
The problem is the following:
If the set is a vector space over $\mathbb R$, write true and explain why each of the following ten properties hold. If the set is not a vector space over $\mathbb R$, write false and identify all of the properties I-X which do not hold.

(a) $\{(x, y) ∈ \mathbb R^2 \:|\: 3x + 2y = 0 \}$ over $\mathbb R$ with the usual definitions of vector addition and scalar multiplication in $\mathbb R^2$

Now, I know what Vector Space over $\mathbb R$ is. I understand the question. There are 10 conditions that I must prove in order to prove that the given is a Vector Space over $\mathbb R$. 
I just don't know where to begin. I understand that if for every $x,y$ in $V$, if $x+y$ exists in $V$, then one of the ten conditions is satisfied. How do I approach this with $3x + 2y = 0$? 
A better question is what the statement $\{(x, y) ∈ \mathbb R^2 \:|\: 3x + 2y = 0 \}$ means. I understand it's all the $x$ and $y$ pairs which satisfy $3x + 2y = 0$, but am I supposed to break the $\mathbb R$ set in all of the possible combinations? For instance, $x = -2$, and $y = 3$ satisfies the function, but $x = 1$, $y = 1$ doesn't. So I certainly can't say that for every $x,y$ in $V$, $3x + 2y$ is in $V$.
I don't know where to begin proving. If I can only get a start somewhere, I would be able to do handle these problems well as I have a general knowledge of what to do, just not how to begin. 
Thanks
Edit 1: Updated the question so it's easy to understand what it is asking.
 A: Consider an element of this set, the element satisfies two conditions. The first is that it must be an element of $\mathbb{R}^2$ and the second is that the components of this vector must satisfy
$$ ax + by = 0 $$
However these two conditions can be rephrased into something clearer. If we were to solve $y$, so the second component of the vector, from the above equation we would find that
$$y = - \frac{a}{b}x$$
So the first condition we stated, namely that the element must be in $\mathbb{R}^2$ is too broad, as there is a fixed condition for what the second component of the vector can be! From these observations we can show the following equality
$$ \{ (x,y) \in \mathbb{R}^2 : ax + by = 0\} = \{ (x, -\frac{a}{b}x) : x \in \mathbb{R}\}$$
As can be seen the second component of the vector is tied to the first component by the relation we derived above. Now the axioms we are interested in showing are considerably easier!
Using the above definition of the set I'll prove that the set is closed under addition. Consider any two elements of this set. The element will be of the form
$$ (v, - \frac{3}{2}v), \ (t,  - \frac{3}{2}t)$$
for some arbitrary $v,t \in \mathbb{R}$. Now consider the sum of these two elements
$$ (v, - \frac{3}{2}v) + (t,  - \frac{3}{2}t) = (v + t, -\frac{3}{2}(v + t))$$
Well $v + t \in \mathbb{R}$ so if we denote $u = v + t$ then the sum of these two vectors gives us a vector
$$ (u, - \frac{3}{2}u)$$
and if you look at the definition of the set above, this element must belong to the set! 
Can you take it from here?
A: Hint:
You can write your set as the vector family
$$(t,-1.5t),\, t\in\Bbb R$$
What can you say about it?
