Đáp án:
$A'\left( {2,3} \right)$; $B'\left( {12,1} \right)$; $C'\left( {6, - 1} \right)$; $D'\left( {0,9} \right)$; $O'\left( {6,9} \right)$
Giải thích các bước giải:
Ta có:
$I\left( {2,3} \right)$
$\begin{array}{l}
+ )A\left( {2,3} \right) \Rightarrow A \equiv I\\
{V_{\left( {I, - 2} \right)}}A = A' \Leftrightarrow \overrightarrow {IA'} = - 2\overrightarrow {IA} = \overrightarrow 0 \Leftrightarrow A' \equiv A\\
\Rightarrow A'\left( {2,3} \right)\\
+ )B\left( { - 3,4} \right) \Rightarrow \overrightarrow {IB} = \left( { - 5,1} \right)\\
{V_{\left( {I, - 2} \right)}}B = B' \Leftrightarrow \overrightarrow {IB'} = - 2\overrightarrow {IB} = \left( {10, - 2} \right)\\
\Rightarrow B'\left( {12,1} \right)\\
+ )C\left( {0,5} \right) \Rightarrow \overrightarrow {IC} = \left( { - 2,2} \right)\\
{V_{\left( {I, - 2} \right)}}C = C' \Leftrightarrow \overrightarrow {IC'} = - 2\overrightarrow {IC} = \left( {4, - 4} \right)\\
\Rightarrow C'\left( {6, - 1} \right)\\
+ )D\left( {3,0} \right) \Rightarrow \overrightarrow {ID} = \left( {1, - 3} \right)\\
{V_{\left( {I, - 2} \right)}}D = D' \Leftrightarrow \overrightarrow {ID'} = - 2\overrightarrow {ID} = \left( { - 2,6} \right)\\
\Rightarrow D'\left( {0,9} \right)\\
+ )O\left( {0,0} \right) \Rightarrow \overrightarrow {IO} = \left( { - 2, - 3} \right)\\
{V_{\left( {I, - 2} \right)}}O = O' \Leftrightarrow \overrightarrow {IO'} = - 2\overrightarrow {IO} = \left( {4,6} \right)\\
\Rightarrow O'\left( {6,9} \right)
\end{array}$
Vậy $A'\left( {2,3} \right)$; $B'\left( {12,1} \right)$; $C'\left( {6, - 1} \right)$; $D'\left( {0,9} \right)$; $O'\left( {6,9} \right)$
