I know we can have a vertical line , something like $(\pi, y)$ for $y$ in some interval, or a horizontal line, something like $(x, \pi)$ for $x$ in some interval, that contains no points $(q_1, q_2)$ s.t. $q_1, q_2 \in \mathbb{Q}$.
I believe that it is possible to have a (continuous, straight) diagonal line segment that has no rational points (mostly because I am unable to prove otherwise!).
Are there any examples of a line like this?
On
You need to take two $\mathbb Q$ linearly independent irrational numbers in $\mathbb R$, $x$ and $y$ (I think $\pi$ and $e$ would do the trick), and look at something like $t(x,y)+(a,b)$ where $a$ and $b$ are rational.
Since $\mathbb Q$ is a field (you can add, subtract, multiply AND divide rational numbers), you can think of $\mathbb R$ as a $\mathbb Q$ vector space! It's infinite dimensional, but there are still bases. There are certainly numbers which are linearaly independent, i.e. there is $x$ and $y$ in $\mathbb R$ such that there are no non-zero rational numbers $a,b$ such that $ax + by = 0$, which has to be the case if you are going to get a rational point on $t(x,y)$.
Here's a very simple example. Let $\alpha$ be any irrational number. Then the line consisting of points of the form $(t,t+\alpha)$ contains no rational points. Indeed, if both $t$ and $t+\alpha$ were rational, then $(t+\alpha)-t=\alpha$ would be rational as well, which is a contradiction.
More generally, any non-vertical line is the graph of a function $f(t)=at+b$ for some $a,b\in\mathbb{R}$. If such a graph contains a rational point, that means there are rational numbers $q$ and $t$ such that $q=at+b$, or $b=q-at$. For any fixed $a$, there are only countably many numbers of the form $q-at$ for $q,t\in\mathbb{Q}$, so for all but countably many choices of $b$, the graph of $f(t)=at+b$ contains no rational points.