Đáp án:
$A'\left( { - 2;0} \right),B'\left( {1;1} \right),C'\left( { - 3;3} \right)$
Giải thích các bước giải:
Gọi \(C\left( {a;a} \right) \in d:x - y = 0\)\( \Rightarrow \overrightarrow {AB} = \left( {2; - 6} \right),\overrightarrow {AC} = \left( {a;a - 4} \right)\)
Tam giác \(ABC\) vuông tại A nên \(\overrightarrow {AB} .\overrightarrow {AC} = 0 \Leftrightarrow 2a - 6\left( {a - 4} \right) = 0\) \( \Leftrightarrow - 4a + 24 = 0 \Leftrightarrow a = 6\)
\( \Rightarrow C\left( {6;6} \right)\).
\(\begin{array}{l}A'' = {V_{\left( {O,\dfrac{1}{2}} \right)}}\left( A \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{A''}} = \dfrac{1}{2}.{x_A} = 0\\{y_{A''}} = \dfrac{1}{2}.{y_A} = 2\end{array} \right.\\B'' = {V_{\left( {O,\dfrac{1}{2}} \right)}}\left( B \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{B''}} = \dfrac{1}{2}.{x_B} = 1\\{y_{B''}} = \dfrac{1}{2}.{y_B} = - 1\end{array} \right.\\C'' = {V_{\left( {O,\dfrac{1}{2}} \right)}}\left( C \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{C''}} = \dfrac{1}{2}.{x_C} = 3\\{y_{C''}} = \dfrac{1}{2}.{y_C} = 3\end{array} \right.\\A' = {Q_{\left( {O,{{90}^0}} \right)}}\left( {A''} \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{A'}} = - {y_{A''}} = - 2\\{y_{A'}} = {x_{A''}} = 0\end{array} \right. \Rightarrow A'\left( { - 2;0} \right)\\B' = {Q_{\left( {O,{{90}^0}} \right)}}\left( {B''} \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{B'}} = - {y_{B''}} = 1\\{y_{B'}} = {x_{B''}} = 1\end{array} \right. \Rightarrow B'\left( {1;1} \right)\\C' = {Q_{\left( {O,{{90}^0}} \right)}}\left( {C''} \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{C'}} = - {y_{C''}} = - 3\\{y_{C'}} = {x_{C''}} = 3\end{array} \right. \Rightarrow C'\left( { - 3;3} \right)\end{array}\)
