Absolute Value equations in 2 variables ($|x-y|+|x+y|=\sqrt5$)

Question

I'm struggling a lot figuring out how to sketch the following $$|x-y|+|x+y|=\sqrt5$$

This is what I know

for$|x-y|$

• $x-y \geq 0$ then $y \leq x$ so $|x-y|= x-y$
• $x-y < 0$ then $y >x$ so $|x-y|= -(x-y)$

for$|x+y|$

• $x+y \geq 0$ then $y \geq -x$ so $|x+y|= x+y$
• $x+y < 0$ then $y <-x$ so $|x+y|= -(x+y)$

but even after knowing this, I still don't know how to sketch it. The result should look something like this:

If you could help me it would be very much appreciated.

Answer 1

$|x-y|+|x+y|=\sqrt5$

We have four possible cases -

$(i) ~ ~ x-y \geq 0, x + y \geq 0$

which is equivalent to $ ~ -x \leq y \leq x, x \geq 0$

That leads to $x = \frac{\sqrt5}{2}$

Similarly,

$(ii) ~ ~ x-y \geq 0, x + y \leq 0$

leads to $ ~ y = - \frac{\sqrt5}{2}$

$(iii) ~ ~ x-y \leq 0, x + y \geq 0$

leads to $y = \frac{\sqrt5}{2}$

$(iv) ~ ~ x-y \leq 0, x + y \leq 0$

leads to $ ~ x = - \frac{\sqrt5}{2}$

Answer 2

You can totally avoid the sign discussion by using this definition of the absolute value: $$|u|=\max(u,-u)$$

Along with this property of the maximum:

$$\max(a,b)+\max(c,d)=\max(a+c,a+d,b+c,b+d)$$

• $|x-y|=\max(x-y,y-x)$
• $|x+y|=\max(x+y,-x-y)$
• $|x-y|+|x+y|=\max(2x,-2y,2y,-2x)=2\max(|x|,|y|)=\sqrt{5}$

We get that $$\max(|x|,|y|)=\frac {\sqrt{5}}2$$ and it is then straightforward to figure out the square shape.