This is, essentially, a geometrically rendered version of [the question](https://mathoverflow.net/q/315973/9924) I asked a week ago, with the emphases slightly shifted; it seems more natural and appealing (to me, at least) in this form. 


Let $p\ge 3$ be a prime number. Suppose we are given $N$ lines $l\_1,\dotsc,l\_N\subset\mathbb F\_p^2$, and we want to translate them to get new lines $l\_1',\dotsc,l\_N'$ (which are either parallel, or identical to the original lines) so as to have the whole vector space $\mathbb F\_p^2$ covered by these new lines: $l\_1'\cup\dotsb\cup l\_N'=\mathbb F\_p^2$. This can be impossible if $N\le2(p-1)$, as it follows by considering the system of $p-1$ "vertical" and $p-1$ "horizontal" lines. Is this always possible if $N\ge 2p-1$?


More generally, given $(p-1)n+1$ affine hyperplanes in $\mathbb F\_p^n$, can one always translate them so that the resulting translates cover the whole space $\mathbb F\_p^n$?



Covering the finite plane with lines
Let $C>0$ be any fixed number. Take $p-3$ horizontal lines and $p-b$ vertical lines where $p\gg b\gg C$. If we want to stay within $2p+C-3$ lines, we should be able to cover some $3\times b$ rectangular configuration by at most $b+C$ lines of any prescribed slopes $a\_1,\dots, a\_{b+C}\ne 0$. Notice that we have $3b$ points to cover and each line can cover at most $3$ points, so we can afford only $3C$ lines that cover $2$ points or fewer. Thus some $b-2C$ lines should pass through $3$ points. 


Let $x\_1,x\_2,\dots,x\_b$ be the base of our $3\times b$ configuration and $u,v,w$ be its "vertical side". Then for each slanted line coming through $3$ points, we have some triple $i,j,k$ such that 
$$
(w-u)(x\_j-x\_i)=(v-u)(x\_k-x\_i) \tag {$\*$}
$$ 
and the slope of the corresponding line is determined by that triple. Notice also that those triples cover at least $b-6C$ indices $1,\dots,b$ (the indices not taken by the exceptional $3C$ lines). Thus we can choose $\frac b3-2C$ linearly independent equations of the type ($\*$) (just take an equation including some index not used yet every time until you run out of them). There are some $K(b)$ possible arrangements of those equations and $p^3$ choices of $u,v,w$, so we see that we can have at most $K(b)p^{3+\frac 23b+2C}$ arrangements that are coverable by $b+C$ slanted lines in principle and all slopes (and even lines) except $3C$ are determined by $x\_j$ and the ($\*$)-equations up to the order. That results in the bound $K(b)(b+C)!p^{3+\frac 23b+5C}$ for the number of choices of $b+C$ slopes that can be used to cover *any* $3\times b$ configuration, which falls short of $(p-1)^{b+C}$ available choices if $b\gg C$ and $p>p(b)$.


If you do it carefully, you get the lower bound of the type $N(p)\ge 2p+p^\alpha$ for large $p$ with some $\alpha\in(0,1)$, but it is still only a rather pitiful improvement of the trivial lower bound $2p-1$, so I didn't try to be very precise in the estimates.


