Note : definitely, not HW, simply personal interest in rotations and linear transformations (cf. my previous questions)
The expression $ x \cos(R)+ y \sin(R)= d$ ( with $d\geq 0$ ) is the equation of a vertical straight line $ x=d$ rotated $R$ radians counterclockwise ( about the origin $(0,0)$) .
Desmos construction : https://www.desmos.com/calculator/g2dps3fccs
Note : in case $d\gt 0$ , this line " slides" on a circle of equation $ x^2 + y^2 = d^2$ ; if $d=0$ , the line turns around the origin.
Given the " rotational" equation $ x cos(R)+ y sin(R)= d$ , one can find the cartesian one by solving for $y$ , thus obtaining :
$$ y = - \frac {\cos(R)} { \sin(R)} x + \frac {d} { \sin(R)} $$.
My question is how to go the other direction : given the equation $ y = mx +b$ , finding the original distance $d$ and the rotation angle $R$ in order to rewrite the equation in the form
$$ x \cos(R)+ y \sin(R)= d$$
or, equivalently, in order to determine of which vertical line the given straight line is the image ( and under which rotation).
You can compare $y=mx+b$ with the "rotational" equation $y = -\cot{R} \cdot x +\frac{d}{\sin{R}}$, to see that
$$m = -\cot{R}$$ $$b = \frac{d}{\sin{R}}$$
The first equation can be used to obtain $R$ from $m$, and then the second allows finding $b$ from $d$ and $R$.