# wizard1's question at Yahoo! Answers (Isometry)

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Here is the question:

Let H = {(x,y) ∈ ℝ^2 : y > 0}, with the usual Euclidean distance. Let T: H→H be the translation map: T(x,y) = (x, y+1) for all (x,y) ∈ H. Verify that T is an isometry of H.
Thanks!
Here is a link to the question:

Verify that T is an isometry of H....? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello wizard1,

If $(x,y)\in H$ then, $y>0$ wich implies $y+1>0$. As a consequence, the map $T:H\to H$ is well defined. Now, for all $(x,y)$ and $(x',y')$ points of $H$: \begin{aligned}d[T(x,y),T(x',y')]&=d[(x,y+1),(x',y'+1)]\\&=\sqrt{(x'-x)^2+(y'+1-x'-1)^2}\\&=\sqrt{(x'-x)^2+(y'-x')^2}\\&=d[(x,y),(x',y')]\end{aligned} That is, $T$ is an isometry.

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