If $||x_0-x_1||=||y_0-y_1||$, there exists an affine isometry such that: $f(x_0)=y_0 , f(x_1)=y_1$

Question

We define:

Affine transfomation: $f:\mathbb{R}^n\to\mathbb{R}^n$ as follows: $f(x)=Ax+b$ as $A\in \mathbb{R}_{nxn}$ and $b\in\mathbb{R}^n$.

Isometry: $\forall x,y\in\mathbb{R}^n: ||f(x)-f(y)||=||x-y||$ as $||\cdot||$ denotes the euclideaמ norm.

Given $x_0,x_1,y_0,y_1 \in \mathbb{R}^n$ staisfying $||x_0-x_1||=||y_0-y_1||$.

Prove: There exists an affine isometry $f:\mathbb{R}^n\to\mathbb{R}^n$ such that:

$f(x_0)=y_0 , f(x_1)=y_1$

I have already showed that an affine transformation is an isometry iff $A$ is orthogonal.

How can I deduce the above claim?

Any help would be appreciated.

Answer 1

1) In the trivial case where $x_{0}=x_{1}$ and $y_{0}=y_{1}$, a translation over the vector $\vec{x_{0}y_{0}}$ does the job.

2) Now assume $x_{0}\neq x_{1}$ and $y_{0}\neq y_{1}$. Let $v_{1}:=\vec{x_{0}x_{1}}/\|\vec{x_{0}x_{1}}\|$ and extend it to an orthonormal basis $\{v_{1},\ldots,v_{n}\}$ of $\mathbb{R}^{n}$. Similarly, let $w_{1}:=\vec{y_{0}y_{1}}/\|\vec{y_{0}y_{1}}\|$ and extend to an orthonormal basis $\{w_{1},\ldots,w_{n}\}$ of $\mathbb{R}^{n}$. Let $A$ be the unique linear transformation of $\mathbb{R}^{n}$ satisfying \begin{cases} A(v_{1})=w_{1}\\ \hspace{1.3cm}\vdots\\ A(v_{n})=w_{n} \end{cases} Put $b:=y_{0}-A(x_{0})$. Now define $$f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}:x\mapsto A(x)+b.$$ I will leave it to you to check that $A$ is orthogonal, and that $f(x_{0})=y_{0}$ and $f(x_{1})=y_{1}$.

Answer 2

Let us consider the vectors $x_1-x_0$ and $y_1-y_0$. Since they have the same norm there exists an isometry $A$ (actually infinitely many if $n>1$) of $\mathbb{R}^n$ such that $(y_1-y_0)=A(x_1-x_0)$. Then, to find (one of) the affine transformation you're after, just put $b=y_0-Ax_0$. This works because $$Ax_0+b=Ax_0-Ax_0+y_0=y_0$$ and $$Ax_1+b=A(x_1-x_0)+y_0=(y_1-y_0)+y_0=y_1$$