\(\overrightarrow{AB}=\left(2;-6\right)\) ; \(\overrightarrow{BC}=\left(2;6\right)\)
Do \(\frac{2}{2}\ne\frac{-6}{6}\Rightarrow\overrightarrow{AB}\) và \(\overrightarrow{BC}\) ko cùng phương hay A; B; C ko thẳng hàng
\(\left\{{}\begin{matrix}x_M=\frac{x_A+x_B}{2}=\frac{-1+1}{2}=0\\y_M=\frac{y_A+y_B}{2}=\frac{4-2}{2}=1\end{matrix}\right.\) \(\Rightarrow M\left(0;1\right)\)
\(\left\{{}\begin{matrix}x_G=\frac{x_A+x_B+x_C}{3}=1\\y_G=\frac{y_A+y_B+y_C}{3}=2\end{matrix}\right.\) \(\Rightarrow G\left(1;2\right)\)
Do D đối xứng A qua G \(\Rightarrow G\) là trung điểm AD
\(\Rightarrow\left\{{}\begin{matrix}x_G=\frac{x_A+x_D}{2}\\y_G=\frac{y_A+y_D}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_D=2x_G-x_A=3\\y_D=2y_G-y_A=0\end{matrix}\right.\) \(\Rightarrow D\left(3;0\right)\)