Is it possible to write an absolute value equation for straight lines, where the symmetry axis is an arbitrary line, in closed form?

Question

Absolute value equations for straight lines where the symmetry axis is either horizontal ($y=y_s$) or vertical ($x=x_s$) can be written in a closed-form manner:

For instance, $|x - 2| + y = 1$ yields the following graph:

while $|2y| + x = 1$ yields:

Is it possible to do this reflection over an arbitrary axis and still be able to write down the equation in closed form? (Thinking out loud: perhaps a rotation transformation can be applied to an equation with a horizontal or vertical symmetry axis in order to obtain the correct angle?)

I ultimately want to write an inequality that shades in one "side" of the kinked line (see below), but I need this inequality to be defined in a single equation (cannot be specified in a piecewise manner, for instance).

$|x - 2| + y \leq 1$ yields:

Thank you very much!

Answer 1

Polar equations says yes!

$x=r\cos(\theta)$ and $y=r\sin(\theta)$. Let's just do some quick graphs centered at $(0,0)$.

$$y=|x|$$

Turn it into polar form:

$$r\sin(\theta)=|r\cos(\theta)|$$

We want to rotate it, say, $\alpha$ degrees, so

$$r\sin(\theta+\alpha)=|r\cos(\theta+\alpha)|$$

Expand with trig identities:

$$r\cos(\alpha)\sin(\theta)+r\sin(\alpha)\cos(\theta)=|r\cos(\alpha)\cos(\theta)-r\sin(\alpha)\sin(\theta)|$$

Return it back to rectangular form,

$$\cos(\alpha)y+\sin(\alpha)x=|\cos(\alpha)x-\sin(\alpha)y|$$

And if $c=\cos(\alpha)$ and $s=\sin(\alpha)$, then

$$cy+sx=|cx-sy|$$