# symmetry with respect to the x-axis and the y-axis

Please help me prove that if a graph is symmetric with respect to the x-axis and to the y-axis, then it is symmetric with respect to the origin.

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aLok: you seem to want to delete this, but it would be both a pity and a little wasteful of Arturo's effort writing a very nice answer (which is quite well-regarded by the community with iots 12 upvotes and all!) Why do you want to delete this? – Mariano Suárez-Alvarez Aug 31 '11 at 19:46

A graph is symmetric about the $x$-axis if and only if whenever $(a,b)$ is in the graph, so is $(a,-b)$.

A graph is symmetric about the $y$-axis if and only if whenever $(a,b)$ is in the graph, so is $(-a,b)$.

A graph is symmetric about the origin if and only if whenever $(a,b)$ is in the graph, so is $(-a,-b)$.

Say you have a point $(a,b)$ on the graph. Can you show (say, in a couple of steps), that symmetry about $x$ and symmetry about $y$, together, imply that $(-a,-b)$ has to be in the graph as well?

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yes.I'am look for a more formal and rigorous proof – alok Aug 24 '11 at 21:26
"More formal rigorous proof" than the one Arturo gave to you? :-? – a.r. Aug 24 '11 at 21:33
I was wondering if there are other ways of proving it. – alok Aug 24 '11 at 22:00
@pencil: I described what is, once you fill in the blank left by the final question, a formal and rigorous proof of the fact in question. Yes, there are other ways of proving it, but if your question is "of all the possible ways of proving it, which is the one that I'm (or person X) is thinking of?", then that's not a real question. – Arturo Magidin Aug 25 '11 at 0:53

Take a point $a$ in the first quadrant (without loss of generality).

Draw a line $l_1$from the origin to $a$ and let $\theta$ be the angle formed by $l_1$ and the x axis.

Reflect $a$ about the $x$ axis and call that point $b$.

Draw the line $l_2$ from the origin to $b$ and call $\beta$ the angle formed by $l_2$ and the $x$ axis, and call $\rho$ the angle formed by $l_2$ and the $y$ axis.

Then $\theta = \beta$ and $l_1 = l_2$.

Now reflect $b$ about the $y$ axis and call that point $c$.

Draw $l_3$ from the origin to $c$ and call $\phi$ the angle formed by $l_3$ and the $y$ axis.

Then $l_2 = l_3$ and therefore $l_3 = l_1$.

Since the $x$ and $y$ axes are orthogonal, $\theta$ and $\beta$ are the complements of $\phi$ and $\rho$, therefore $\theta + \beta + \phi + \rho = 180$ degrees, and $l_1 + l_3$ is the diameter of the circle with the origin as center.

Therefore $c$ is symmetric to $a$ with respect to the origin.

I strongly suggest drawing this out.

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