One can proceed by cases:
Suppose $b = 0$, so that we have the line $V = ax + c$. Note that to actually have a line in the first place, this entails $a \neq 0$.
Consider the homomorphism $\phi: \Bbb R[x] \to \Bbb R$ given by $\phi(f(x)) = f\left(\dfrac{-c}{a}\right)$.
Show this induces a homomorphism $\psi:\Bbb R[x,y] \to \Bbb R[y]$, by considering $\Bbb R[x,y]$ as $(\Bbb R[x])[y]$, and writing:
$\psi(g(x,y)) = \psi(f_0(x) + f_1(x)y + \cdots + f_n(x)y^n) = \phi(f_0(x)) + \phi(f_1(x))y + \cdots + \phi(f_n(x))y^n$.
Use the fact that $\phi$ is onto $\Bbb R$ (why?) to conclude $\psi$ is onto $\Bbb R[y]$.
Now show the kernel of $\psi$ consists precisely of those elements of $\Bbb R[x,y]$ that, when viewed as elements of $(\Bbb R[x])[y]$ (strictly speaking, we are employing a "unnamed" isomorphism between $\Bbb R[x,y]$ and $(\Bbb R[x])[y]$), have coefficients that lie in the kernel of $\phi$, which is to say they all have a common factor of $\left(x + \dfrac{c}{a}\right)$.
Conclude that the kernel of $\psi$ is indeed the ideal generated by $ax + c$.
Case 2 is somewhat easier (and shows why we treated case 1 separately): Suppose $b \neq 0$, and consider the homomorphism:
$\phi: \Bbb R[x,y] \to \Bbb R[x]$ given by $\phi(g(x,y)) = g\left(x,\dfrac{-a}{b}x - \dfrac{c}{b}\right)$
Show that $g \in \text{ker }\phi \iff (ax + by + c)|g$, and that $g$ maps constant polynomials to their constant terms (that is, acts as the identity on $\Bbb R$).
Note: although it "looks" different, we are actually using the same trick of considering $\Bbb R[x,y]$ as $(\Bbb R[x])[y]$, and "evaluating" $g(x,y)$ at the ring element $f(x) = \dfrac{-a}{b}x - \dfrac{c}{b}$ in $\Bbb R[x]$.
Some random thoughts: what is a line, really? Let's be a *bit* more specific, and consider a line embedded in a Euclidean plane. Normally, we think of it as points satisfying an equation:
$L = \{(x,y) \in \Bbb R^2: Ax + By + C = 0\}$.
But this equation depends on our "coordinate system", we have to "locate" the $x$ and $y$ axes to come up with the numbers $A,B,C$. That works fine for a *vector space*, where we have a distinguished origin. But we actually think of the plane as being "undifferentiated", any part of it looks like any other. In an affine space, there isn't any special point of origin. So what remains the same? The relationship between the points is "linear", that is, the locus of the line represents the zero-set of a linear polynomial in $\Bbb R[x,y]$. In other words we have moved away from any *specific* $A,B,C$, and have characterized a line as being determined by the DEGREE of a polynomial in two variables. This characterization is now independent of a coordinate-system, as it should be, in our understanding of just "an abstract line" (which ought to be a line no matter what coordinates we are using).
I urge you to consider what "affine varieties" we obtain by considering the zero-sets of degree TWO polynomials in $\Bbb R[x,y]$ (hint: you already know these-the parabola is one example).
Degree three and higher (what are commonly called "algebraic curves") polynomial zero-sets for $\Bbb R[x,y]$ in the plane start getting a tad complicated. Go figure.
Mathbalarka: if I've made any silly mistakes, please point them out-my mind seems to be slipping, lately :).