Let $\mathbb{F}_q$ be a finite field of order $q$, where $q$ is a prime power. Show that the set $\mathcal{L} = \{\{(x,y) : ax + by = c\} : a,b,c \in \mathbb{F}_q \text{, and at least one of a,b is non-zero}\}$ has size $q^2+q$.
Please give comments on the following approach. Let $g \in \mathbb{F}_q, g \neq 0$. If $m,n,r \in [1,q]$ are integers, then we can represent $a,b,c$ as $a = g^m, b = g^n \text{ and } c = g^r$. Since $g^q = 0$, I tried doing a case analysis for $m = q$ and $m \neq q$. But this would double count the multiples of the equations $ax + by = c$. I guess there are some other properties of finite field that I'm missing?
There are $q^3$ triples $(a,b,c)\in{\mathbb F}_q^3$. Among them the $q$ triples $(0,0,c)$ are forbidden. Therefore we have $q^3-q$ equations of the form $$ax+by=c\tag{1}$$ defining a line $\ell\subset{\mathbb F}_q^3$. Now multiplication of the coefficient triple $(a,b,c)$ in $(1)$ with some $\lambda\in{\mathbb F}_q^*$ does not change $\ell$. It follows that each $\ell$ occurs $q-1$ times among the $q^3-q$ equations $(1)$. Therefore there are $${q^3-q\over q-1}=q^2+q$$ different lines in ${\mathbb F}_q^3$.