I have two well-behaved curves in the plane $\gamma_1,\gamma_2:\mathbb{R}\rightarrow \mathbb{R}^2$. Call their coordinates $\gamma_i = \langle f_i, g_i\rangle$.
Given any two points on these curves, I can draw the line between them. By definition, it has slope $$A(u,v) \equiv \frac{g_1(u) - g_2(v)}{f_1(u) - f_2(v)}$$ and y-intercept $$B(u,v) = \frac{f_1(u)g_2(v) - f_2(v)g_1(u)}{f_1(u)-f_2(v)}$$.
Because these curves are so well-behaved, I can solve for the curves given the functional form of $A(u,v)$ and $B(u,v)$. The solution is \begin{align*} f_1 &= -\frac{\partial_v B}{\partial_v A}\\\\ g_1 &= B - A\frac{\partial_v B}{\partial_v A} \end{align*}
The surprise is that this functional form for finding the x- and y- coordinates of each curve looks similar to when you have an equation for a line $y = ax + b$, which you solve for the x-intercept $x_0 = -b/a$ and then plug into the original equation $y = ax_0 + b$.
I am curious about the analogy between these two forms. More precisely, I am wondering if this correspondence represents some kind of isomorphism between the algebraic structure of the lines and the differential structure of the slope-intercept fields $A$ and $B$.
The farthest I've gotten is to note that the nondegeneracy condition for the two curves (i.e. that they are smooth, locally injective, and there exists a unique line between any point on one curve and any point on the other), means that the jacobian of $A,B$ is nonzero and so $\langle A,B\rangle$ represent a kind of coordinate transform.
