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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.  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 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.

See more: Words Ending In Ch - List Of Words Ending With 'Ch'

I strongly suggest illustration this out.

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answered May 30 "15 at 15:51 \$endgroup\$

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