Trong mặt phẳng tọa độ $Oxy$, phép đối xứng trục biến điểm $A\left( {2;1} \right)$ thành $A'\left( {2;5} \right)$ có trục đối xứng là:

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{{36}} + \dfrac{{k2\pi }}{3}\\x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)$

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3}\\x = \dfrac{{5\pi }}{{36}} + \dfrac{{k\pi }}{3}\,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)$

$\left[ \begin{array}{l}x = \dfrac{\pi }{{36}} + \dfrac{{k2\pi }}{3}\\x =  - \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)$

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{{36}} + k2\pi \\x = \dfrac{{5\pi }}{{36}} + k2\pi \,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)$

$\min y = 1;\max y = 4033.$

$\min y =  - 1;\max y = 4033.$

$\min y = 1;\max y = 4022.$

$\min y =  - 1;\max y = 4022.$

$1$     

$0$      

$3$

$2$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{8} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{8} + k\pi ,k \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{4} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}$ 

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{3} + k2\pi ,\dfrac{{5\pi }}{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

${\rm{D}} = \left\{ {\dfrac{\pi }{3} + k2\pi ,\dfrac{{5\pi }}{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{{5\pi }}{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

$\overrightarrow {MM'}  = \overrightarrow {NN'}$ 

$\overrightarrow {MM'}  = \overrightarrow {N'N}$ 

$\overrightarrow {MN'}  = \overrightarrow {NM'}$ 

$\overrightarrow {MN}  = \overrightarrow u$