Determine the intersection of a set of boundary lines I have a set of boundary line pairs. I am interested in the intersection of the regions defined by the pairs of boundary lines.
The diagram below illustrates. I am trying to solve for the yellow region. The black lines represent (in this case) a set of two of boundary line pairs. Note that the number of boundary line pairs is one or more.

At the moment I use a brute force approach: for every discrete point in my $2D$ space (image space in this case), for every pair of boundary lines, I determine whether the point exists between them using a simple side-of-line test. I test whether the point exists inside of all pairs of boundary lines.
The above approach is painfully slow, and it bugs me that it is very inefficient. My question is, is it possible to mathematically determine the region of a $2D$ space that is the union of a collection of parallel boundary lines?
 A: At a practical level, you have the advantage that the intersection is always convex. For this, you can use something like the Liang Barsky or Sutherland-Hodgman algorithms for "clipping". I think that SH is better tuned to what you want to do. (You'll be clipping your image-plane rectangle by the interiors of the parallel-line pairs, which are almost rectangles, except that their ends are at $\pm \infty$. If your image plane lies in the range $-A \le x, y \le A$, then you can safely truncate your infinite strips to ones with coordinates that range from, say, $-5A$ to $5A$, so you end up with finite rectangles to feed into the SH algorithm.
The downside is that for $n$ line-pairs, applying SH repeatedly may still involve $O(n^2)$ work, even if the final answer is very simple (e.g., it turns out to be just a rectangle or parallelogram). To avoid this $n^2$ work requires anticipating which line to clip in which order, and that may well be difficult. 
A: You describe testing whether a "point exists betweeen [a pair of boundary lines] using a simple side-of-line test."
Of course this requires a specification not only of a line:
$$ \text{ line } L:  ax + by = c $$
where not both $a,b$ are zero, but also of one side or the other of the line on which the points should lie.  This amounts to an inequality in place of the equality for the line itself, e.g.
$$ \text{ half plane } L^\ge: ax + by \ge c $$
We can of course switch the direction of the inequality to designate the "other side" of the line, or we could just as well keep the same direction but switch all the signs of $a,b,c$.  Note that if we changed the weak inequality $\ge$ to a strict inequality $\gt$, this would have the effect of excluding the points which actually lie exactly on the line $L$.
Now consider a pair of lines and what it means for a point to be "between" them.  If the lines were parallel (and distinct) it would be unambiguous what non-empty set would be described by this, but for a pair of non-parallel lines, it is necessary to specify which side of each line we want.  Thus there are four possible convex "regions" which are intersections of the respective two pairs of half-planes:

To say you want the intersection of $n$ regions formed by pairs of boundary lines (intersecting half planes) is the same as wanting the intersection of $2n$ half planes.  The intersection is known as the feasibility region in linear programming, a topic of mathematics concerned with solutions to systems of linear inequalities.
Depending on your application, an efficient way of determining the feasible region (which may be empty or a nonempty convex subset; if nonempty then bounded or unbounded) may be worth implementing.  All I know about your situation is that the problem is limited to two-dimensions, which is a significant simplification.
With more information I might be able to recommend an implementation approach.
A: Elaborating on @hardmath's answer, first re-phrase the problem as the intersection of $2N$ half-planes.
One can solve for the intersection of these 2N half-planes using linear programming. There are also bespoke algorithms for the intersection-of-half-planes problem that may be faster, depending on your application.
For example, here is an approach by Preparata and Muller for solving the intersection of $n$ half-spaces with $n\log(n)$ time complexity. There are other variants (e.g. Wu, Ji, and Chen), but they all have the same complexity.
These lecture notes by Dave Mount are especially useful for understanding the the math and geometry underlying these algorithms. These notes show how to construct the (convex) intersection set as the intersection of a (convex) upper and lower envelope.
