Showing that a given map is indeed an isometry.

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I have two problems here and my attempts to solve them. I need some help seeing if I'm doing them correctly or not.

1 ) Let $R : \Bbb R → \Bbb R$ be the reflection map $R(x) = a − x$ $\forall x ∈ \Bbb R$ and for some fixed real number $a$. Verify that $R$ is an isometry of the real line $\Bbb R$.

For this I said $d(x,y) = |x-y|$ and $d(R(x), R(y)) =|(a-x)-(a-y)|= |y-x| = |x-y|$ So since $d(x,y) = d(R(x), R(y))$, this is an isometry.

2 ) Let $(a, b)$ be any point in $\Bbb R^2$. Let $T : \Bbb R^2 → \Bbb R^2$ be the translation map$: T(x, y)=(x + a, y + b)$ $\forall (x, y) ∈ \Bbb R^2$. Verify that $T$ is an isometry of $\Bbb R^2$.

For this I said $d((x,y),(r,s)) = \sqrt{(x-r)^2+(y-s)^2}$ and $d(T(x,y),T(r,s)) = \sqrt{((x+a)-(r+a))^2+((y+b)-(s+b))^2} = \sqrt{(x-r)^2+(y-s)^2}$. So since $d((x,y),(r,s)) = d(T(x,y),T(r,s))$, this is an isometry as well.

Am I correct in my thought process? Any help would be appreciated.