Đáp án:
$\left( {{d_1}} \right):2x - 3y + 1 = 0$
Giải thích các bước giải:
Lấy ${A_2}\left( {2; - 1} \right) \in \left( {{d_2}} \right)$
Ta có:
$\begin{array}{l}
{T_{\overrightarrow a = \left( {1; - 2} \right)}}{A_1} = {A_2}\left( {2; - 1} \right) \Rightarrow {A_1}\left( {1;1} \right)\\
{T_{\overrightarrow a = \left( {1; - 2} \right)}}\left( {{d_1}} \right) = \left( {{d_2}} \right) \Rightarrow \left( {{d_2}} \right)//\left( {{d_1}} \right)\\
\Rightarrow A \in \left( {{d_1}} \right);\overrightarrow {{n_{{d_1}}}} = \overrightarrow {{n_{{d_2}}}} \\
\Rightarrow \left( {{d_1}} \right):2\left( {x - 1} \right) - 3\left( {y - 1} \right) = 0\\
\Rightarrow \left( {{d_1}} \right):2x - 3y + 1 = 0
\end{array}$
Vậy $\left( {{d_1}} \right):2x - 3y + 1 = 0$
