Show that a $2D$ region is symmetric around the line $y=x$ or $y=-x$ I need to show that the following region is symmetric around the line $y=x$ and $y=-x$:
$$\{(x,y)|x^2+y^2+xy\le0.5\}$$
How can I do it algebraically (without drawing the region)?
 A: Let $f(x, y) = x^2 + y^2 +xy$.
Let a general point $P = (x, y)$. Let point $P$ have the mirror image $P' = (x', y')$ about a given line $l$.
A sufficient condition for symmetry of $f$ about $l$ is that $f(x, y) = f(x', y')$ for all possible $P$.
To gain an intuitive understanding of why this works, think of an easier example in a single variable, for instance, the parabola $f(x) =(x-a)(x-b)$ and how you would show it's symmetric about the line $x= m = \frac{a+b}2$. Here $x' = 2m - x = a+b-x \iff x = a+b - x'$. $f(x) = (x-a)(x-b) = (a+b-x'-a) (a+b-x'-b) = (b-x')(a-x') = (x'-a) (x'-b) =f(x')$ and symmetry about the line $x=m$ is proven.
Now, for your question, note that the reflection of $P = (x, y)$ about the line $y=x$ is $P' = (x', y') = (y, x)$. You need to show that $f(x, y) = f(y, x) = f(x', y')$ (i.e. swapping $x$ and $y$ in the expression defining the region leaves its value invariant).
Similarly, the reflection of $P = (x, y)$ about the line $y=-x$ is $P' = (x', y') = (-y, -x)$ and again you need to show that so you need to show that $f(x, y) = f(y, x) = f(x', y')$, (i.e. replacing $x$ with $-y$ while replacing $y$ with $-x$ leaves the expression invariant).
The actual algebra is completely trivial.
By the way, the constant term on the right hand side and the nature of the inequality are not relevant here. If every point and its mirror image return exactly the same value from the variable expression, then symmetry is assured.
