Assume I have an implicit function $F(x,y,z)=0$ which can be expressed as a family of lines, i.e. $$ F(x,y,z)=0 \quad\text{if and only if}\quad y=m(z)x+b(z) $$ for functions $m$ and $b$.
Is it possible, for general $m$ and $b$, to put this relation into determinantal form as understood in nomography?
On
Here is an answer without determinant.
Indeed, one can use - if it exists - the envelope curve of all these straight lines. This envelope is classicaly obtained by considering the linear system
$$\begin{cases}y&=&m(z)x+b(z) & (a)\\ 0&=&m'(z)x+b'(z) & (a')\\ \end{cases} \tag{1}$$
(where (a') is obtained by differentiation of (a) with respect to parameter $z$).
From (1), one obtains a parametric representation of the envelope curve by "extracting" $x$ and then $y$ as functions of $z$ alone:
$$\begin{cases}x&=&-b'(z)/m'(z)\\ y&=&(m'(z)b(z)+m(z)b'(z))/m'(z)\\ \end{cases} \tag{2}$$
In this way, after having placed "tick marks" on this envelope, one gets all the $x$ and $y$ satisfying relationship $y=m(z)x+b(z).$
On
The equation can also be nomographed directly using the Type 10 nomographs implemented in PyNomo.
After reading up on Warmus' procedure it became clear that an answer to my question is $$ \det \left|\begin{array}{ccc} 1 & x & 0 \\ 0 & y & 1 \\ -m(z) & b(z) & 1\end{array}\right|. $$