Understanding the graph of $|x|-x = |y|- y$

Question

What does the graph for $\left|x\right|-x=\left|y\right|-y$ look like for $x, y \ge0$

When considering absolute values of $x$ and $y$ there are 4 cases:

1. $x\lt 0 \; y\lt 0$
2. $x\ge 0 \; y\lt 0$
3. $x \lt 0 \; y\ge 0$
4. $x\ge0, \; y\ge 0$

Solutions for the first three ones are:

1. $x = y; \; x,y < 0$
2. $y = 0; \; x\ge0, y<0$
3. $x = 0; \; x<0, y\ge0$

But for $x, y \ge 0$ the equation holds for every single value of $x$ and $y$. Does that mean that the whole plane $x, y \ge 0$ is part of the graph? If so then what tools could I use to see that? Desmos and W|A are not showing that the plane is part of the graph.

Here is how Mathematica plots the function:

Answer 1

$$|x|-x=|y|-y\tag 1$$

Case $x> 0$ and $y>0$ : $$|x|-x=0=|y|-y=0 \qquad\text{Eq.}(1)\text{ is true any }x>0,y>0.$$

Case $x<0$ and $y<0$ : $$|x|-x=-x-x=-2x=|y|-y=-y-y=-2y$$ $$y(x)=x\qquad\text{ any }x<0.$$

Case $x>0$ and $y<0$ : $$|x|-x=0=-y-y=-2y \qquad\text{no solution since }y<0\neq 0.$$

Case $x<0$ and $y>0$ : $$|x|-x=-x-x=-2x=y-y=0\qquad\text{no solution since }x<0\neq 0.$$

Case $x=0$ and $y>0$ : $$|x|-x=0=|y|-y=0 \qquad\text{Eq.}(1)\text{ is true any }y>0.$$

Case $x=0$ and $y<0$ : $$|x|-x=0=|y|-y=-2y \qquad\text{no solution since }y<0\neq 0.$$

Case $x>0$ and $y=0$ : $$|x|-x=0=|y|-y=0 \qquad\text{Eq.}(1)\text{ is true any }x>0.$$

Case $x<0$ and $y=0$ : $$|x|-x=-2x=|y|-y=0 \qquad\text{no solution since }x<0\neq 0.$$

Case $x=0$ and $y=0$ : Eq.$(1)$ is true.

On figure below, the red points satisfy Eq.$(1)$.

Answer 2

The program Graph found here is somewhat out of date, but it does plot your question correctly.