What is the equation for a line in a family ruling a rectangle, given the rectangle's dimensions and a point on the line inside the rectangle? I have questions about the following ruling of rectangle $ABCD$ with top and bottom sides, $AB$ and $DC$, being collinear with $y=1$ and $y=−1$, respectively, and the midpoint of each lies on $x=0$. The ruling is defined as the sweep of a line through the rectangle as it follows this path: given a line collinear with side $AB$, rotate counterclockwise around vertex $B$ until it intersects with the opposite vertex $D$ (i.e., is collinear with diagonal $BD$), then rotate it clockwise around vertex $D$ until it is collinear with side $DC$.
There are an infinite number of families of lines that fit these criteria, one family per rectangle with a unique ratio of adjacent side lengths; for all families, it follows that: a) every point inside the rectangle lies on exactly one of the lines, so no two lines intersect, b) every line that is not a side of ABCD has a positive slope, and c) every line has a unique intercept.
I'm trying to find a general equation for lines in such a family, which takes as variables the ratio of one side of the rectangle to its adjacent side (to identify the family) and the xy coordinates of a point inside the rectangle, and returns the slope of the line in that family through the given point.
I suspect what I'm describing is equivalent to a rectangle cut out of the side of a single-sheet hyperboloid or some other quadric, but even so, I don't know how to go from the equation for the hyperboloid to the equation I want, other than the diagonal of the rectangle would be an asymptote of the hyperbola.
Am I right that this equation must exist, and, if so, what is it or how can I find it? Finally, is the equation associated with a particular quadric or other surface?
 A: Here's a re-statement of my re-statement of the problem (approved by OP in a comment, but with an orientation fix and an explicit vertical dimension for sanity-checking purposes):


Consider an axis-aligned, clockwise-oriented rectangle $\square ABCD$ with
$$A=(−r,s) \qquad B=(r,s) \qquad C=(r,-s) \qquad D=(−r,-s)$$
and consider a point $P=(p,q)\in \square ABCD$. We seek a function $f(p,q,r,s)$
such that
$$f(p,q,r,s) = \begin{cases} \text{equation of $\overleftrightarrow{PB}$}, & P\in\triangle ABD\\[4pt]
\text{equation of $\overleftrightarrow{PD}$}, & P\in\triangle CDB \end{cases} \tag{1}$$

The line equations are easy enough to generate:
$$\begin{align}
\overleftrightarrow{PB}:&\quad (s-q) x - (r-p) y = ps - qr \\
\overleftrightarrow{PD}:&\quad (s+q) x - (r+p) y = ps - qr
\end{align} \tag{2}$$
The cases are determined by whether $P$ lies above or below the diagonal $\overline{BD}$. The equation of that diagonal is $sx-ry=0$, so the condition corresponds to whether $(x,y)\to(p,q)$ makes $sx-ry$ negative (for "above") or positive (for "below"). With an explicit zero case, we have
$$f(p,q,r,s) = \begin{cases} 
(s-q) x - (r-p) y = ps - qr, & ps < qr \\[4pt]
sx - ry = 0, & ps = qr \\[4pt]
(s+q) x - (r+p) y = ps - qr, & ps > qr \end{cases} \tag{3}$$
Defining $\lambda := \operatorname{sgn}(ps-qr)$, we can unify the above into

$$f(p,q,r,s) \quad\to\quad (s+\lambda q) x - (r+\lambda p) y = ps - qr \tag{$\star$}$$

or this symmetric-looking alternative

$$f(p,q,r,s) \quad\to\quad (s+\lambda q) (x-p) = (r+\lambda p) (y-q) \tag{$\star\star$}$$

Substituting $s=1$ for OP's particular needs is left as an exercise to the reader. :)
Here's the function in action (with $P$ not constrained to lie within the rectangle):

