Symmetry in the line $x=y$

In the plane, why does the symmetry in the line $y=x$ sends the point of coordinates $(x,y)$ into the point of coordinates $(y,x)$ ?

P.S. : does someone have a proof without linear algebra ?

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Here's a little geometric proof using basic algebra.

We are interested in a point $(a,b)$. Clearly the reflection of $(a,b)$ through $y=x$ is going to be on the line through $(a,b)$ perpendicular to $y=x$. Since the perpendicular has slope -1, the equation of that line is: $y=-(x-a)+b$. (I'm omitting the case when $a=b$ because in that case it is clear that it is fixed by the reflection.)

Clearly $(b,a)$ is also on that line, but if it truly is going to be the reflection, it must be the same distance as $(a,b)$ is from the intersection of the two lines.

Basic algebra will tell you they intersect at $((a+b)/2,(a+b)/2)$. Now, if you check the distance from this point to $(a,b)$ and to $(b,a)$, you will find they both have the same distance: $|a-b|$. Thus the two points are reflections of each other.

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Thank you very much ! – Charles Jan 18 '13 at 19:00

It works fine for the special cases $(1,1)$ and $(1,-1)$, hence for all their linear combinations.

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In the plane, why does the symmetry in the line $y=x$ send the point of coordinates $(x,y)$ into the point of coordinates $(y,x)$ ? Does someone have a proof without linear algebra ?

Note that given the line of symmetry $\;y = x \iff x = y,\;$, it is fitting that for the "flip" corresponding to a symmetric mapping about this line both of the following hold:

Every $\;x$-value is mapped to $\;y\;$ while every $\;y$-value is mapped to $x$: i.e. $\;(x, y) \to (y, x)$.

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