a: A(-2;1); B(3;2); C(-1;4)
\(AB=\sqrt{\left(3+2\right)^2+\left(2-1\right)^2}=\sqrt{5^2+1^2}=\sqrt{26}\)
\(AC=\sqrt{\left(-1+2\right)^2+\left(4-1\right)^2}=\sqrt{1^2+3^2}=\sqrt{10}\)
\(BC=\sqrt{\left(-1-3\right)^2+\left(4-2\right)^2}=\sqrt{2^2+\left(-4\right)^2}=2\sqrt{5}\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{26+10-20}{2\cdot\sqrt{26}\cdot\sqrt{10}}=\dfrac{4}{\sqrt{65}}\)
=>\(sinBAC=\sqrt{1-cos^2BAC}=\dfrac{7}{\sqrt{65}}\)
Diện tích tam giác ABC là:
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\dfrac{7}{\sqrt{65}}\cdot\sqrt{26}\cdot\sqrt{10}=7\)
b: ADBC là hình thoi
=>AB\(\perp\)DC và \(\overrightarrow{AD}=\overrightarrow{CB}\)
=>\(\left\{{}\begin{matrix}\overrightarrow{AD}=\overrightarrow{CB}\\\overrightarrow{AB}\cdot\overrightarrow{DC}=0\end{matrix}\right.\)
\(\overrightarrow{AD}=\left(x+2;y-1\right);\overrightarrow{CB}=\left(4;-2\right)\)
\(\overrightarrow{AB}=\left(5;1\right);\overrightarrow{DC}=\left(-1-x;4-y\right)\)
Do đó, ta có: \(\left\{{}\begin{matrix}x+2=4\\y-1=-2\\5\left(-x-1\right)+1\left(4-y\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=-1\\5\left(-2-1\right)+1\left(4+1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=-1\\5\cdot\left(-3\right)+1\cdot5=0\left(sai\right)\end{matrix}\right.\)
vậy: Không có điểm D nào thỏa mãn