K đối xứng B qua G \(\Rightarrow\overrightarrow{BG}=\overrightarrow{GK}=\frac{1}{2}\overrightarrow{BK}\)
Theo t/c trọng tâm:
\(\overrightarrow{BG}=\frac{1}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}=\frac{1}{3}\overrightarrow{BA}+\frac{1}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AK}=\overrightarrow{AB}+\overrightarrow{BK}=\overrightarrow{AB}+2\overrightarrow{BG}=\overrightarrow{AB}+2\left(-\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
\(=\overrightarrow{AB}-\frac{4}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}=-\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\frac{1}{3}\\y=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow3x+3y=1\)
