Giải thích các bước giải:
Ta có:
$A\left( {2,1} \right);B\left( {2, - 1} \right);C\left( { - 2, - 3} \right)$
a) Gọi $D\left( {x,y} \right)$
Để $ABCD$ là hình bình hành
$\begin{array}{l}
\Leftrightarrow \overrightarrow {BA} = \overrightarrow {CD} \\
\Leftrightarrow \left\{ \begin{array}{l}
x - \left( { - 2} \right) = 2 - 2\\
y - \left( { - 3} \right) = 1 - \left( { - 1} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 2 = 0\\
y + 3 = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 2\\
y = - 1
\end{array} \right.
\end{array}$
Vậy $D\left( { - 2, - 1} \right)$
b) Gọi $M\left( {x,y} \right)$
$ \Rightarrow \overrightarrow {MA} = \left( {2 - x,1 - y} \right);\overrightarrow {MB} = \left( {2 - x, - 1 - y} \right);\overrightarrow {MC} = \left( { - 2 - x, - 3 - y} \right)$
$ \Rightarrow 3\overrightarrow {MA} + \overrightarrow {MB} - 2\overrightarrow {MC} = \left( {12 - 2x,8 - 2y} \right)$
Để $3\overrightarrow {MA} + \overrightarrow {MB} - 2\overrightarrow {MC} = \overrightarrow 0 $
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
12 - 2x = 0\\
8 - 2y = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 6\\
y = 4
\end{array} \right.
\end{array}$
Vậy $M\left( {6,4} \right)$
c) Gọi $E\left( {x,0} \right)$
$\begin{array}{l}
{S_{ABE}} = \frac{1}{2}AB.d\left( {E,AB} \right)\\
\Leftrightarrow \frac{1}{2}.2.\left| {x - 2} \right| = 24\\
\Leftrightarrow \left| {x - 2} \right| = 24\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 24\\
x - 2 = - 24
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 26\\
x = - 22
\end{array} \right.
\end{array}$
Vậy $E\left( {26,0} \right)$ hoặc $E\left( { - 22,0} \right)$
