Đáp án:
$\begin{array}{l}
a)D\left( {x;y} \right) \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {AB} = \left( {4;1} \right)\\
\overrightarrow {DC} = \left( {2 - x; - 4 - y} \right)
\end{array} \right.\\
ABCD\,la\,hinh\,binh\,hanh\\
\Rightarrow \overrightarrow {AB} = \overrightarrow {DC} \\
\Rightarrow \left( {4;1} \right) = \left( {2 - x; - 4 - y} \right)\\
\Rightarrow \left\{ \begin{array}{l}
2 - x = 4\\
- 4 - y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - 2\\
y = - 5
\end{array} \right.\\
\Rightarrow D\left( { - 2; - 5} \right)\\
c)AB = CD = \sqrt {{4^2} + {1^2}} = \sqrt {17} \\
AD = BC = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( { - 4 - 2} \right)}^2}} = \sqrt {37} \\
\Rightarrow {P_{ABCD}} = 2\sqrt {17} + 2\sqrt {37}
\end{array}$
Gọi pt đường thẳng đi qua A và B là y=ax+b
Nên ta có hệ pt:
$\begin{array}{l}
\left\{ \begin{array}{l}
1 = - 3a + b\\
2 = a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = \frac{1}{4}\\
b = \frac{7}{4}
\end{array} \right.\\
\Rightarrow AB:y = \frac{1}{4}x + \frac{7}{4}hay\,:x - 4y + 7 = 0\\
\Rightarrow {d_{D - AB}} = \frac{{\left| { - 2 - 4.\left( { - 5} \right) + 7} \right|}}{{\sqrt {{1^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{25}}{{\sqrt {17} }}\\
\Rightarrow {S_{ABCD}} = AB.{d_{D - AB}} = \sqrt {17} .\frac{{25}}{{\sqrt {17} }} = 25
\end{array}$
