\(\overrightarrow{AB}=\left(2;1\right);\overrightarrow{CB}=\left(3;2\right);\overrightarrow{CA}=\left(1;1\right)\)
\(\Rightarrow AB=\sqrt{5};BC=\sqrt{13};AC=\sqrt{2}\)
\(\Rightarrow AB+BC+AC=\sqrt{2}+\sqrt{5}+\sqrt{13}\)
\(cos\widehat{BCA}=\frac{BC^2+AC^2-AB^2}{2BC.AC}=\frac{5\sqrt{26}}{26}\)
ABCD là hbh \(\Rightarrow\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}1-x_D=2\\-y_D=1\end{matrix}\right.\) \(\Rightarrow D\left(-1;-1\right)\)
Do \(AH\perp BC\Rightarrow AH\) nhận \(\left(3;2\right)\) là 1 vtpt
Phương trình AH:
\(3\left(x-2\right)+2\left(y-1\right)=0\Leftrightarrow3x+2y-8=0\)
Phương trình BC:
\(2\left(x-1\right)-3y=0\Leftrightarrow2x-3y-2=0\)
Tọa độ H là nghiệm \(\left\{{}\begin{matrix}3x+2y-8=0\\2x-3y-2=0\end{matrix}\right.\) \(\Rightarrow H\left(\frac{28}{13};\frac{10}{13}\right)\)
