Đáp án: A(-4;12); B( 6;4)
Giải thích các bước giải:
$\begin{array}{l}
M\left( {2;0} \right) \Rightarrow \left\{ \begin{array}{l}
2 = \frac{{{x_B} + {x_C}}}{2}\\
0 = \frac{{{y_B} + {y_C}}}{2}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
4 = {x_B} + \left( { - 2} \right)\\
0 = {y_B} + \left( { - 4} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_B} = 6\\
{y_B} = 4
\end{array} \right. \Rightarrow B\left( {6;4} \right)\\
G\left( {0;4} \right) \Rightarrow \left\{ \begin{array}{l}
0 = \frac{{{x_A} + {x_B} + {x_C}}}{3}\\
4 = \frac{{{y_A} + {y_B} + {y_C}}}{3}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
0 = {x_A} + 6 - 2\\
12 = {y_A} + 4 - 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_A} = - 4\\
{y_A} = 12
\end{array} \right. \Rightarrow A\left( { - 4;12} \right)
\end{array}$
