Let $q$ be an odd prime power and $\mathbb{F}_q$ be the finite field of $q$ element. For $a, b, c, d$ are non-zero in $\mathbb{F}_q$ with $a + b = c + d \neq 0$. How many quadruples $(x, y, z, t) \in \mathbb{F}_q^4$ such that $x, y, z, t$ are distinct and $$ ax + by = cz + dt \quad \text{ and } \quad ax^2 + by^2 = cz^2 + dt^2. $$
In my attempt, WLOG we can assume that $a + b = c + d = 1$. Thus, the equations become $ax + (1-a)y = cz + (1-c)t$ and $ax^2 + (1-a)y^2 = cz^2 + (1-c)t^2$. Hence $$ a(a-1)(x-y)^2 = c(c-1)(z-t)^2. $$ But from here, I don't know what should be going next. Perhaps, the system of equations has no solution. I can prove that this system has no solution when in the case $a = c$ or $a = d$.