\(\left( d \right):\,\,2x+y+1=0;\,\,\left( \Delta ...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử \(\overrightarrow{u}=\left( a;b \right)\) .

* Giả sử \(M\left( x;y \right)\in \left( d \right).\,\,{{T}_{\overrightarrow{u}}}:\,\,M\mapsto M'\left( x';y' \right).\) Ta có :

\(\left\{ \begin{array}{l}
x' = x + a\\
y' = y + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = x' - a\\
y = y' - b
\end{array} \right. \Rightarrow M\left( {x' - a;y' - b} \right)\)

* \(M\in \left( d \right)\Rightarrow 2\left( x'-a \right)+\left( y'-b \right)+1=0\)

\(\Rightarrow 2x'+y'-2a-b+1=0\)

Mặt khác \(\left( d' \right):\,\,2x+y-6=0\)

\(\Rightarrow -2a-b+1=-6\Rightarrow -2a-b+7=0\,\,\left( 1 \right)\)

* Giả sử \(N\left( x;y \right)\in \left( \Delta  \right).\,\,{{T}_{\overrightarrow{u}}}:\,\,N\mapsto N'\left( x';y' \right).\) Ta có :

\(\left\{ \begin{array}{l}
x' = x + a\\
y' = y + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = x' - a\\
y = y' - b
\end{array} \right. \Rightarrow N\left( {x' - a;y' - b} \right)\)

* \(N\in \left( \Delta  \right)\Rightarrow \left( x'-a \right)+\left( y'-b \right)+15=0\)

\(\Rightarrow x'+y'-a-b+15=0\)

Mặt khác \(\left( \Delta ' \right):\,\,x+y+10=0\Rightarrow -a-b+15=10\Rightarrow -a-b+5=0\,\,\,\left( 2 \right)\)

* Giải hệ 

\(\left\{ \begin{array}{l}
\left( 1 \right)\\
\left( 2 \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
- 2a - b + 7 = 0\\
- a - b + 5 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 2\\
b = 3
\end{array} \right.\)

\(\Rightarrow \overrightarrow{u}=\left( 2;3 \right)\).

Chọn A