Câu 13:
\(A' = {T_{\overrightarrow v }}\left( A \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{A'}} = 3 + 2 = 5\\{y_{A'}} = 5 + 1 = 6\end{array} \right. \Rightarrow A'\left( {5;6} \right)\)
\(B\left( { - 2; - 1} \right) \in d,B' = {T_{\overrightarrow v }}\left( B \right) \Rightarrow \left\{ \begin{array}{l}{x_{B'}} = - 2 + 2 = 0\\{y_{B'}} = - 1 + 1 = 0\end{array} \right. \Rightarrow B'\left( {0;0} \right)\)
Gọi \(d':2x + 3y + c = 0\) là ảnh của \(d\) qua \({T_{\overrightarrow v }}\).
Khi đó \(B' \in d' \Leftrightarrow 2.0 + 3.0 + c = 0 \Leftrightarrow c = 0\)
Vậy \(d':2x + 3y = 0\).
Câu 14:
Lấy \(A\left( {0;4} \right),B\left( {2;0} \right) \in d\).
\(\begin{array}{l}A' = {Q_{\left( {O,{{90}^0}} \right)}}\left( A \right) \Rightarrow A'\left( { - 4;0} \right)\\B' = {Q_{\left( {O,{{90}^0}} \right)}}\left( B \right) \Rightarrow B'\left( {0;2} \right)\end{array}\)
\(\begin{array}{l}A'' = {V_{\left( {O,3} \right)}}\left( {A'} \right) \Rightarrow A''\left( { - 12;0} \right)\\B'' = {V_{\left( {O;3} \right)}}\left( {B'} \right) \Rightarrow B''\left( {0;6} \right)\end{array}\)
\( \Rightarrow d''\) đi qua \(A'',B''\) nên có phương trình \(\dfrac{x}{{ - 12}} + \dfrac{y}{6} = 1\)
