Nomogram for family of lines

Question

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?

Answer 1

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 $y=m(z)x+b(z).$

Answer 2

After reading up on Warmus' procedure it became clear that an answer to my question is $$ \det \left|\begin{array}{ccc} 1 & x & 0 \\ 0 & y & 1 \\ -m(z) & b(z) & 1\end{array}\right|. $$

Answer 3

The equation can also be nomographed directly using the Type 10 nomographs implemented in PyNomo.