Determine two curves from the slope between them Let $\gamma_1, \gamma_2 : \mathbb{R} \rightarrow \mathbb{R}^2$ be two non-intersecting smooth curves.  We do not know what these curves are, but we do have for each pair of inputs $\langle u, v\rangle$ the slope and y-intercept of the line joining $\gamma_1(u)$ to $\gamma_2(v)$. Denote the slope by $A(u,v)$ and y-intercept by $B(u,v)$.
Is it possible to compute some $\gamma_1, \gamma_2$ which fit the constraints enforced by $A$ and $B$? I realize that there might be several solutions, but I am only looking for a method which yields one, where we are given that one exists. 
In my solution attempts, I have attempted to fix arbitrary boundary conditions such as $\vec{p} = \gamma_1(u_0)$, $\vec{q} = \gamma_2(v_0)$, then attempted to find differential equations for $\gamma_1$ and $\gamma_2$ by considering the lines through the image of $u$, $v$, $u+\delta u$, $v+\delta v$.

If it helps, the curves $\gamma_i$ are well-behaved in various senses. For example, they never self-intersect, and the Jacobian $\partial(A,B)/\partial(u,v)$ is never zero.  
 A: Here is a partial solution:
Label the coordinates $\gamma_1(u) = \langle f(u), g(u)\rangle$. By definition, we have that the point-slope equation holds: 
$$\forall u, v :\;\;\;g(u) = A(u,v) f(u) + B(u,v)$$
which implies that for particular values $v_0, v_1$ and all values $u$:
\begin{align*}
g(u) &= A(u, v_0) f(u) + B(u, v_0)\\
g(u) &= A(u, v_1) f(u) + B(u, v_1)\\
\end{align*}
Hence, equating the two right hand sides, we obtain
$$\left[A(u, v_0) - A(u, v_1)\right] f(u) = -\left[B(u,v_0)-B(u,v_1)\right] $$
Because this equation holds for all $u$, $v_0$, $v_1$, we can in particular search for appropriate $v_0(u)$ and $v_1(u)$ for each $u$ to ensure that $A(u,v_0) - A(u, v_1) \neq 0$. I believe the condition $\partial(A,B)/\partial(u,v) \neq 0$ everywhere is enough to guarantee the existence of such points (?).
In this case, we can solve directly:
\begin{align}
f(u) &= \frac{-\left[B(u,v_0) - B(u,v_1)\right]}{\left[A(u,v_0)-A(u,v_1)\right]}\\
g(u) &= A(u,v_0) f(u) + B(u, v_0)\\
\end{align}
And similarly for $\gamma_2(v) = \langle f_2(v), g_2(v)\rangle$.
Hence we can solve for $\gamma_1$ and $\gamma_2$, though I am surprised that the solutions are apparently unique; I suspect there is more to this problem.

Edit: Interestingly, the equation where we solve for $f(u)$ suggests that we can divide numerator and denominator by $(v_0-v_1)$ then (because $A$ and $B$ are presumed smooth) take the limit as $v_1 \rightarrow v_0$. In such a way, we obtain the result that
$$\forall v_0, \qquad f(u) = -\frac{\partial_v B(u, v_0)}{\partial_v A(u,v_0)}$$
