a/ Để tứ giác ABCD là hbh
\(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\)
\(\Leftrightarrow\left(x_B-x_A;y_B-y_A\right)=\left(x_C-x_D;y_C-y_D\right)\)
\(\Leftrightarrow\left(3;-7\right)=\left(2-x_D;1-y_D\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2-x_D=3\\1-y_D=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_D=-1\\y_D=8\end{matrix}\right.\Rightarrow D\left(-1;8\right)\)
b/ \(x_G=\frac{x_A+x_B+x_C}{3}=\frac{-2+1+2}{3}=\frac{1}{3}\)
\(y_G=\frac{y_A+y_B+y_C}{3}=\frac{4-3+1}{3}=\frac{2}{3}\)
\(\Rightarrow G\left(\frac{1}{3};\frac{2}{3}\right)\)
