a,
\(D\left(x;y\right)\rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;1\right)\\\overrightarrow{DC}=\left(2-x;-4-y\right)\end{matrix}\right.\)
\(\Rightarrow\) ABCD là hình bình hành
\(\overrightarrow{AB}=\overrightarrow{DC}\)
\(\rightarrow\left(4;1\right)=\left(2-x;-4-y\right)\)
\(\left\{{}\begin{matrix}x=-2\\y=-5\end{matrix}\right.\)
\(\rightarrow D=\left(-2;-5\right)\)
b. \(AB=CD=\sqrt{4^2+1^2=\sqrt{17}}\)
\(AD=BC=\sqrt{\left(2-1\right)^2+\left(-4-1\right)^2}=\sqrt{37}\)
\(\rightarrow P_{ABCD}=2\sqrt{17}+2\sqrt{37}\)
Gọi pt đường thẳng đi qua A và B là y=ax+b
Nên ta có hệ pt:
\(\left\{{}\begin{matrix}1=-3a+b\\2=a+b\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=\frac{1}{4}\\b=\frac{7}{4}\end{matrix}\right.\)
\(\rightarrow AB:y=\frac{1}{4}x+\frac{7}{4}hay:x-47+7=0\)
\(d_{D-AB}=\frac{|2-4.\left(-5\right)+7|}{\sqrt{1^2+\left(-4\right)^2}}=\frac{25}{\sqrt{17}}\)
\(S_{ABCD}=AB.d_{D-AB}=\sqrt{17}.\frac{25}{\sqrt{17}}=25\)
