For a $2 \times 2$ integer matrix with determinant $-1$, i.e. a hyperbolic toral automorphism, it's quite easy to write down an explicit formula for the eigenlines (spans of eigenvectors). Using that formula, it's also not hard to show that the angle between the two eigenlines can be chosen arbitrarily close to any given angle, and that (without fixing the angle) one of the lines can be chosen to have slope arbitrarily close to any given slope.
However, it's not obvious to me that these two choices can be made simultaneously. That is: let $\ell_1, \ell_2$ be two lines through the origin in $\mathbb{R}^2$. Let $\varepsilon > 0$ and pick your favourite metric on the space of pairs of lines through the origin. Is there a $2 \times 2$ integer matrix with determinant $-1$ whose eigenlines are $\varepsilon$-close to $\ell_1, \ell_2$?
I think I have an answer. Please let me know if there are any mistakes.
Let $(a,b;c,d)$ be a matrix and set $s=a+d$, $t=a-d$, $u=b+c$, $v=b-c$. And $s^2 - t^2 - (u^2 - v^2) = 4$. The values $(u+v, -t \pm \sqrt{s^2+1})$ yield points on the eigenlines of $(a,b;c,d)$.
Now given two distinct lines with rational coefficients say $y = p_+x$ and $y = p_-x$ (with $p_+>p_-$). Then choose $k$ large enough so that $s=(p_+-p_-)k$ and $t = (p_++p_-)k$ are integral (and any quantity mentioned hereafter). One can verify that $-t\pm s$ lie on the lines for $x=2k$. The task is now to choose $u+v = 2k$ such that $s^2 - t^2 - (u^2 - v^2) = 4$. A quick computation shows that this can be done in such a way such that $u,v$ have the same even/odd parity. Then the mapping $s,t,u,v\to a,b,c,d$ given in the obvious way yields a matrix with determinant $-1$ and whose eigenlines approximate the rational lines above. The full generality follows by noting that rational lines can approximate any line to arbitrary degree.