Let $a_k,b_k \in \mathbb R$ for $k \in [4] := \{1,2,3,4\}$. Define $f_k:\mathbb R \to \mathbb R$ by $f(x) := a_kx+b_k$. For any $i \in [4]$, define $A_i := \{x \in \mathbb R \mid f_i(x) = \max_{k \in [4]} f_{k}(x)\}$. It is clear that $A_i$ is an interval, say with extremities $\alpha_i$ and $\beta_i$ where $-\infty \le \alpha_i \le \beta_i \le \infty$.
Question. What is a closed-form expression for $\alpha_i$ and $\beta_i$ in terms of the $a_k$'s and $b_k$'s.
You need to break it up into different cases.
Let $1\leq i\leq 4$. Then $x\in A_i$ if and only if $(a_i-a_j)x\geq b_j-b_i$ for each $j\neq i$. If $a_i>a_j$, then $x\geq\frac{b_j-b_i}{a_i-a_j}$. If $a_i<a_j$, then $x\leq\frac{b_j-b_i}{a_i-a_j}$. If $a_i=a_j$, then all we know is that $b_i\geq b_j$.
To give one example: If we look at the case where the $a_i$ are distinct (so that there are no parallel lines) then, without loss of generality, we may assume that $a_1<a_2<a_3<a_4$.
Then $x\in A_2$ if and only if $(a_2-a_j)x\geq b_j-b_2$ for each $j\neq 2$. Then $a_2>a_1$, so $x\geq\frac{b_1-b_2}{a_2-a_1}$. Also, $a_2<a_3,a_4$, so $x\leq\frac{b_j-b_2}{a_2-a_j}$ for $j=3,4$. So $$A_2=\left[\frac{b_1-b_2}{a_2-a_1},\min\left\{\frac{b_3-b_2}{a_2-a_3},\frac{b_4-b_2}{a_2-a_4}\right\}\right].$$