> [!NOTE] Lemma (Isometry that fixes $O$ and $e_{1}$ is a reflection in $x$-axis or identity) > Let $f:\mathbb{R}^{2}\to \mathbb{R}^{2}$ be a [[Isometry of The Plane|plane isometry]] such that $f(0,0)=(0,0)$ and $f(1,0)=(1,0).$ Then $f$ is either the [[Identity Function|identity]] or a reflection in the $x$-axis. *Proof*. Let $(s,t)= f(0,1).$ Then $(s,t)$ must lie on the unit circle since it has the same distance from the origin as $(0,1).$ It must also be the same also be the same distance from $(1,0)$ as $(0,1)$ so it lies on the circle centred at $(1,0)$ with radius $2.$ So $(s,t)$ is on both of these circles which means that $s^{2}+t^{2}=1 \quad \text{and} \quad (s-1)^{2}+t^{2}=2$Solving this simultaneously gives $s=0$ and $t= \pm 1.$ Let $\underline{p} =(x,y)\in\mathbb{R}^{2}$ and $\underline{q}=f(p)=(u,v)\in\mathbb{R}^{2}.$ *Case 1*: Suppose $f(0,1)=(0,1).$ The distance of $\underline{p}$ from $(0,0)$ is the same as the distance of $\underline{q}$ from $f(0,0)=(0,0).$ Therefore $u^{2}+v^{2}=x^{2}+y^{2} \tag{1}$The distance of $\underline{p}$ from $(1,0)$ is the same as $\underline{q}$ from $f(1,0)=(1,0).$ Therefore $(u-1)^{2}+v^{2}=(x-1)^{2}+y^{2}\tag{2}$Considering also the distance of $\underline{p}$ from $(0,1)$ gives $u^{2}+(v-1)^{2}=x^{2}+(y-1)^{2} \tag{3}$Now $(1)-(3)$ gives $v^{2} - (v-1)^{2}= y^{2} -(y-1)^{2}$hence $v=y.$ So $x=u$ and $\underline{p}=\underline{q}$ thus $f$ is the identity map. *Case 2*: Suppose $f(0,1)=(0,-1).$ As in case 1, we can deduce that $x=u.$ Considering the distance of $\underline{p}$ from $(0,1)$ gives $u^{2}+(v+1)^{2}=x^{2}+(y-1)^{2} \tag{4}$Now $(1)-(4)$ gives $v=-y.$ Therefore $f(x,y)=(x,-y)$ and $f$ is the reflection in the $x$-axis.