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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.
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?
answered Aug 24 “11 at 19:13
Arturo MagidinArturo Magidin
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Take a point $a$ in the first quadrant (without loss of generality).
Draw a line $l_1$from the origin to $a$ and let $ heta$ 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 $eta$ the angle formed by $l_2$ and the $x$ axis, and call $
ho$ the angle formed by $l_2$ and the $y$ axis.
Then $ heta = eta$ 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, $ heta$ and $eta$ are the complements of $phi$ and $
ho$, therefore $ heta + eta + phi +
ho = 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.
answered May 30 “15 at 15:51
jeremy radcliffjeremy radcliff
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