a: ABCD là hình bình hành
=>vecto AB=vecto DC
=>\(\left\{{}\begin{matrix}0-x_D=4-2=2\\-1-y_D=5+3=8\end{matrix}\right.\Leftrightarrow D\left(-2;-9\right)\)
b: \(AB=\sqrt{\left(4-2\right)^2+\left(5+3\right)^2}=2\sqrt{17}\)
\(AC=\sqrt{\left(0-2\right)^2+\left(-1+3\right)^2}=2\sqrt{2}\)
\(BC=\sqrt{\left(0-4\right)^2+\left(-1-5\right)^2}=2\sqrt{13}\)
Xét ΔBAC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{68+8-52}{2\cdot2\sqrt{2}\cdot2\sqrt{17}}=\dfrac{3\sqrt{34}}{34}\)
nên gócA=59 độ
\(sinA=\sqrt{1-\left(\dfrac{3}{\sqrt{34}}\right)^2}=\dfrac{5}{\sqrt{34}}\)
\(S=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot2\sqrt{17}\cdot2\sqrt{2}\cdot\dfrac{3}{\sqrt{34}}=\dfrac{1}{2}\cdot2\cdot2\cdot3=6\)
