Do ABCD là hbh \(\Rightarrow\overrightarrow{AI}=\overrightarrow{IC}=\frac{1}{2}\overrightarrow{AC}\Rightarrow\overrightarrow{AN}=\frac{1}{2}\overrightarrow{AI}=\frac{1}{4}\overrightarrow{AC}\)
Gọi \(A\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AN}=\left(-3-x;2-y\right)\\\overrightarrow{AC}=\left(3-x;-y\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-3-x=\frac{1}{4}\left(3-x\right)\\2-y=\frac{1}{4}\left(-y\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=\frac{8}{3}\end{matrix}\right.\) \(\Rightarrow A\left(-5;\frac{8}{3}\right)\)
G là trọng tâm ABC \(\Rightarrow\left\{{}\begin{matrix}x_B=3x_G-x_A-x_C=5\\y_B=3y_G-y_A-y_C=\frac{1}{3}\end{matrix}\right.\)
ABCD là hbh \(\Rightarrow\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow D...\)
