# Does having two lines of symmetry $y=0$ and $y=-x$ imply that the shape also has lines of symmetry $x=0$ and $y=x$

I have been wondering if a shape/curve that has a line of symmetry along the lines $$y=0$$ and $$y=-x$$ is guaranteed to also have lines of symmetry along the lines $$x=0$$ and $$y=x$$.

My gut feeling tells me that this is true. All the shapes that I can think of satisfy this. But I can't think of a way to prove this.

Also, is this relationship "if and only if" (with a $$\Leftrightarrow$$ symbol) or just "implies" (with a $$\Rightarrow$$ symbol).

I tried looking up similar questions here but I couldn't find any. Maybe I worded it too verbosely.

Let $$(u,v)$$ be a point on the curve. Then:

• symmetry about $$y=0$$ implies that $$(u,-v)$$ is on the curve;
• symmetry about $$y=-x$$ implies that $$(-v,-u)$$ is on the curve.

Now we can apply these rules iteratively:

• reflect $$(u,-v)$$ about $$y=-x$$ to get $$(v,-u)$$
• reflect $$(v,-u)$$ about $$y=0$$ to get $$(v,u)$$

And $$(v,u)$$ is $$(u,v)$$ reflected about $$y=x$$. So $$y=x$$ is indeed a line of symmetry.

Similarly, reflecting $$(-v,-u)$$ about $$y=0$$ then $$y=-x$$ gives $$(-u,v)$$, which is $$(u,v)$$ reflected about $$x=0$$. So $$x=0$$ is also a line of symmetry.

Yes, this is true.

Consider a point $$(x,y)$$ in the $$1^{\text{st}}$$ quadrant. By line of symmetry $$y=0$$, the curve also contains $$(-x,y)$$. Then by the line of symmetry $$y=-x$$, the curve contains $$(-x,-y)$$ (since it contains $$(x,y)$$). Now, the curve must contain $$(-x,y)$$ since the line of symmetry $$y=0.$$

Therefore we have proved the $$4$$ points $$(x,y),(x,-y),(-x,-y),(-x,y)$$ are simultaneously on the curve or off the curve.

Then you can try prove it has lines of symmetry $$x=0$$ and $$y=x$$. $$(*)$$

Comment if you got stuck on how to prove $$(*)$$ or if I have made anything unclear.