The way I solved the problem is to change the equation to $|x+2|=1-|y-3|$, and then square both sides. But I don't think it is the right way to solve the problem. I hope someone can either give me a hint or show me how to solve the problem.
$|x+2|+|y-3|=1$ is an equation for a square. How many units are in the lengths of its diagonals?
The maximum difference of the $x$s comes when $y=3$ so the second absolute value is zero. That means $|x+2|=1$, which gives $x=-3,-1$. One diagonal is from $(-3,3)$ to $(-1,3)$ and is $2$ units long. The other diagonal is along the line $x=-2$.
This is the same as $|x|+|y| = 1$, but shifted so that its center is at $(-2, 3)$.
This is the square with vertices $(1, 0), (0, 1), (-1, 0), (0, -1)$.
The diagonals are $(1, 0)$ to $(-1, 0)$ and $(0, 1)$ to $(0, -1)$ and their lengths are each $2$.
