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


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

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


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answered Aug 24 "11 at 19:13
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Arturo MagidinArturo Magidin
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Take a suggest $a$ in the initially quadrant (without loss of generality).

Draw a line $l_1$from the beginning to $a$ and let $ heta$ be the angle created by $l_1$ and also the x axis.

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

Draw the line $l_2$ from the origin to $b$ and contact $eta$ the angle created by $l_2$ and also the $x$ axis, and call $ ho$ the angle formed by $l_2$ and also the $y$ axis.

Then $ heta = eta$ and $l_1 = l_2$.

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

Draw $l_3$ from the origin to $c$ and also speak to $phi$ the angle created 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$ levels, and $l_1 + l_3$ is the diameter of the circle through the beginning as facility.

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

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I strongly suggest illustration this out.


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answered May 30 "15 at 15:51
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jeremy radcliffjeremy radcliff
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