a/ Theo công thức trọng tâm: \(\left\{{}\begin{matrix}x_G=\frac{x_M+x_N+x_P}{3}=\frac{2}{3}\\y_G=\frac{y_M+y_N+y_P}{3}=-1\\\end{matrix}\right.\) \(\Rightarrow G\left(\frac{2}{3};-1\right)\)
b/ Gọi \(Q\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MN}=\left(0;-2\right)\\\overrightarrow{QP}=\left(-2-x;-3-y\right)\end{matrix}\right.\)
Để MNPQ là hbh \(\Leftrightarrow\overrightarrow{MN}=\overrightarrow{QP}\)
\(\Rightarrow\left\{{}\begin{matrix}-2-x=0\\-3-y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\) \(\Rightarrow Q\left(-2;-1\right)\)
c/ \(\overrightarrow{MQ}=\left(-4;-2\right)\Rightarrow\left|\overrightarrow{MQ}\right|=\sqrt{\left(-4\right)^2+\left(-2\right)^2}=2\sqrt{5}\)
Mà \(MQ=NP\) (tính chất hbh) \(\Rightarrow\left|\overrightarrow{NP}\right|=2\sqrt{5}\)
\(\overrightarrow{NP}=\overrightarrow{MQ}=\left(-4;-2\right)\)
d/ I là trung điểm đường chéo MP
\(\Rightarrow\left\{{}\begin{matrix}x_I=\frac{x_M+x_P}{2}=0\\y_I=\frac{y_M+y_P}{2}=-1\end{matrix}\right.\) \(\Rightarrow I\left(0;-1\right)\)
