\(\overrightarrow{MB}=\left(1-x_M;2-y_M\right)\)
\(\overrightarrow{MC}=\left(4-x_M;1-y_M\right)\)
\(\overrightarrow{AB}=\left(-1;-3\right)\)
Theo đề, ta có:
\(\left\{{}\begin{matrix}1-x_M+3\left(4-x_M\right)=-2\\2-y_M+3\left(1-y_M\right)=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-x_M+12-3x_M=-2\\2-y_M+3-3y_M=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-4x_M=-15\\-4y_M=-11\end{matrix}\right.\Leftrightarrow M\left(\dfrac{15}{4};\dfrac{11}{4}\right)\)