d nhận $
\overrightarrow{\mathbf{AB}}\left({4\mathrm{;}6}\right)
$
làm véc tơ chỉ phương:
PTTS: $
\left\{{\begin{array}{l}
{{1}\mathrm{{+}}{4}{t}}\\
{\mathrm{{-}}{2}\mathrm{{+}}{6}{t}}
\end{array}}\right.
$
Câu tiếp theo là véc tơ AM = vtOA + 2vtMB chứ
$
\overrightarrow{\mathbf{AM}}\hspace{0.33em}\mathrm{{=}}\hspace{0.33em}\overrightarrow{\mathbf{OA}}\mathrm{{+}}{2}\overrightarrow{\mathbf{MB}}
$
Gọi M (x;y) ta có:
$
\begin{array}{l}
{{\mathrm{(}}{x}\mathrm{{-}}{1}{\mathrm{;}}{y}\mathrm{{+}}{2}{\mathrm{)}}}\\
{\overrightarrow{\mathbf{OA}}{\mathrm{(}}\mathrm{{-}}{1}{\mathrm{;}}{2}{\mathrm{)}}}\\
{\overrightarrow{\mathbf{MB}}{\mathrm{(}}{x}\mathrm{{+}}{3}{\mathrm{;}}{y}\mathrm{{-}}{4}{\mathrm{)}}}\\
{\mathrm{\Rightarrow}\hspace{0.33em}{x}\mathrm{{-}}{1}\hspace{0.33em}\mathrm{{=}}\hspace{0.33em}\mathrm{{-}}{1}\hspace{0.33em}\mathrm{{+}}\hspace{0.33em}{2}{\mathrm{(}}{x}\mathrm{{+}}{3}{\mathrm{)}}\mathrm{\Longleftrightarrow}{x}\mathrm{{=}}\mathrm{{-}}{6}}\\
{\mathrm{\Rightarrow}\hspace{0.33em}{y}\mathrm{{+}}{2}\hspace{0.33em}\mathrm{{=}}\hspace{0.33em}{2}\hspace{0.33em}\mathrm{{+}}{2}{\mathrm{(}}{y}\mathrm{{-}}{4}{\mathrm{)}}\hspace{0.33em}\mathrm{\Longleftrightarrow}\hspace{0.33em}{y}\hspace{0.33em}\mathrm{{=}}{\mathrm{10}}}
\end{array}
$
=> M(-6;10)
PTTS: $
\left\{{\begin{array}{l}
{{x}\mathrm{{=}}\mathrm{{-}}{6}\mathrm{{-}}{4}{t}}\\
{{y}\mathrm{{=}}{\mathrm{10}}\mathrm{{+}}{9}{t}}
\end{array}}\right.
$
