Đáp án:
$\begin{array}{l}
a)D\left( {x;y} \right)\\
ABCD\,la\,hình\,bình\,hành\\
\Rightarrow \overrightarrow {AB} = \overrightarrow {DC} \\
\Rightarrow \left( {6;2} \right) = \left( {2 - x;1 - y} \right)\\
\Rightarrow \left\{ \begin{array}{l}
6 = 2 - x\\
2 = 1 - y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - 4\\
y = - 1
\end{array} \right.\\
\Rightarrow D\left( { - 4; - 1} \right)\\
b)G\,là\,trọng\,tâm\,\Delta ABC\\
\Rightarrow \left\{ \begin{array}{l}
{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3}\\
{y_G} = \frac{{{y_A} + {y_B} + {y_C}}}{3}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_G} = \frac{{ - 4 + 2 + 2}}{3}\\
{y_G} = \frac{{1 + 3 + 1}}{3}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 0\\
y = \frac{5}{3}
\end{array} \right.\\
Vậy\,G\left( {0;\frac{5}{3}} \right)
\end{array}$
