This is the continuation of this question: What is the size of the set of lines in a finite field $\mathbb{F}_q$ of order $q$, where $q$ is a prime power?
Now I need to show that exactly $q$ points belong to each line. Let $g \in \mathbb{F}_q, g \neq 0$. If $m,n,r,s,t \in [1,q]$ are integers, then we can represent $a,b,c$ as $a = g^m, b = g^n, c = g^r, x = g^s \text{ and } y = g^t$. Given $a,b,c$, we can then write the equation as $g^mg^s + g^ng^t = g^r$. Since $q$ is a prime power, I deduce that $m + s + n + t \equiv r \text{ mod } q$. Fixing a value of $s$ also fixes the value of $t$, so we have $q$ pairs $(x,y)=(g^s,g^t)$ for each line.
Please give your comment. Also my algebra is pretty rusty, so I think there should be a clearer way to say what I meant. If so please correct my answer.
Embed the affine plane $F_q \times F_q$ in the projective plane where $$(x,y) \mapsto (x:y:1)$$. Any line in the affine plane has the same number of points as any other line and the same is true in the projective plane. The points on the line $x=0$ in the affine plane are in 1-1 correspondence with the points $x=0$ in the projective plane except for the point (0:1:0). Any such point can be written uniquely as (0:b:1) where b is in $F_q$. There are $q$ such points.