Đáp án:
\(d':\,\,3x + y - 15 = 0\)
Giải thích các bước giải:
\(\begin{array}{l}
Lay\,\,M\left( {x;y} \right) \in d\\
Goi\,\,M'\left( {x';y'} \right) = {V_{\left( {O; - 3} \right)}}\left( M \right)\\
\Rightarrow \left\{ \begin{array}{l}
x' = - 3x\\
y' = - 3y
\end{array} \right.\\
Goi\,\,M''\left( {x'';y''} \right) = {T_{\overrightarrow v }}\left( M \right)\\
\Rightarrow \left\{ \begin{array}{l}
x'' = x' + 1\\
y'' = y' - 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x'' = - 3x + 1\\
y'' = - 3y - 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3x = 1 - x''\\
3y = - 3 - y''
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \frac{1}{3}x'' + \frac{1}{3}\\
y = - \frac{1}{3}y'' - 1
\end{array} \right.\\
\Rightarrow M\left( { - \frac{1}{3}x'' + \frac{1}{3}; - \frac{1}{3}y'' - 1} \right)\\
M \in d\\
\Rightarrow 3\left( { - \frac{1}{3}x'' + \frac{1}{3}} \right) - \frac{1}{3}y'' - 1 - 5 = 0\\
\Leftrightarrow - x'' + 1 - \frac{1}{3}y'' - 6 = 0\\
\Leftrightarrow x'' + \frac{1}{3}y'' - 5 = 0\\
\Leftrightarrow 3x'' + y'' - 15 = 0\\
\Rightarrow d':\,\,3x + y - 15 = 0
\end{array}\)
