# determining rotations and reflections in $\mathbb{R}^{2}$

Let $E\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be a rotation about some point $(x_{0},y_{0})$, and let $\theta$ the angle of rotation for $E$. A formula for $E$ can be determined as follows.

First, translate the point $(x_{0},y_{0})$ to $(0,0)$. The map of this translation is $T(x,y)=(x-x_{0},y-y_{0})$.

Next, rotate by $\theta$ about the origin. The map of this rotation is $R(x,y)=(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$.

Finally, translate the point $(0,0)$ back to $(x_{0},y_{0})$. The map of this translation is $T^{-1}(x,y)=(x+x_{0},y+y_{0})$.

The fact that $E=T^{-1}\circ R\circ T$ can be used to obtain a formula for $E$:

$\begin{array}[]{rl}E(x,y)&=(T^{-1}\circ R\circ T)(x,y)\\ &=(T^{-1}\circ R)(x-x_{1},y-y_{0})\\ &=T^{-1}((x-x_{0})\cos\theta-(y-y_{0})\sin\theta,(x-x_{0})\sin\theta+(y-y_{0})% \cos\theta)\\ &=((x-x_{0})\cos\theta-(y-y_{0})\sin\theta+x_{0},(x-x_{0})\sin\theta+(y-y_{0})% \cos\theta+y_{0}).\end{array}$

Let $E\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be a reflection about some line $y=mx+b$. Let $\theta=2\arctan m$. A formula for $E$ can be determined as follows.

First, translate the $y$ intercept $(0,b)$ to $(0,0)$. The map of this translation is $B(x,y)=(x,y-b)$.

Next, reflect about the line $y=mx$. The map of this reflection is $F(x,y)=(x\cos\theta+y\sin\theta,x\sin\theta-y\cos\theta)$.

Finally, translate the point $(0,0)$ back to $(0,b)$. The map of this translation is $B^{-1}(x,y)=(x,y+b)$.

The fact that $E=B^{-1}\circ F\circ B$ can be used to obtain a formula for $E$:

$\begin{array}[]{rl}E(x,y)&=(B^{-1}\circ F\circ B)(x,y)\\ &=(B^{-1}\circ F)(x,y-b)\\ &=B^{-1}(x\cos\theta+(y-b)\sin\theta,x\sin\theta-(y-b)\cos\theta)\\ &=(x\cos\theta+(y-b)\sin\theta,x\sin\theta-(y-b)\cos\theta+b).\end{array}$

Let $E\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be a reflection about some line $x=c$. A formula for $E$ can be determined as follows.

First, translate the $x$ intercept $(c,0)$ to $(0,0)$. The map of this translation is $C(x,y)=(x-c,y)$.

Next, reflect about the line $x=0$. The map of this reflection is $M(x,y)=(-x,y)$.

Finally, translate the point $(0,0)$ back to $(c,0)$. The map of this translation is $C^{-1}(x,y)=(x+c,y)$.

The fact that $E=C^{-1}\circ M\circ C$ can be used to obtain a formula for $E$:

$\begin{array}[]{rl}E(x,y)&=(C^{-1}\circ M\circ C)(x,y)\\ &=(C^{-1}\circ M)(x-c,y)\\ &=C^{-1}(c-x,y)\\ &=(2c-x,y).\end{array}$

Title determining rotations and reflections in $\mathbb{R}^{2}$ DeterminingRotationsAndReflectionsInmathbbR2 2013-03-22 17:14:34 2013-03-22 17:14:34 Wkbj79 (1863) Wkbj79 (1863) 4 Wkbj79 (1863) Derivation msc 51A15 msc 51A10 msc 15A04 Symmetry2