a) Ta có: $\vec{BA}=(1;3)\Rightarrow BA=\sqrt{1^2+3^2}=\sqrt{10}$
$\vec{BC}=(-3;1)\Rightarrow BC=\sqrt{(-3)^2+1^2}=\sqrt{10}$
$\Rightarrow BA=BC\Rightarrow \Delta ABC$ cân đỉnh $B$
Lại có $\vec{BA}.\vec{BC}=1.(-3)+3.1=0$
$\Rightarrow\vec{BA}\bot \vec{BC}\Rightarrow BA\bot BC\Rightarrow \widehat{ABC}=90^o$
$\Rightarrow ABC$ vuông cân đỉnh $B$
b) $S_{ABC}=\dfrac{1}{2}BA.BC=\dfrac{1}{2}\sqrt{10}.\sqrt{10}=5$ (đơn vị diện tích)
$\dfrac{1}{BH^2}=\dfrac{1}{BA^2}+\dfrac{1}{BC^2}=2\dfrac{1}{BA^2}=2.\dfrac{1}{10}=\dfrac{1}{5}$
$\Rightarrow BH=\sqrt5$
c) Áp dụng định lý Pitago vào $\Delta $ vuông $ABC$
$AC^2=BC^2+BA^2=\sqrt{10}^2+\sqrt{10}^2=20$
$\Rightarrow AC=2\sqrt5$
$\Rightarrow P_{ABC}=BA+BC+AC=\sqrt{10}+\sqrt{10}+2\sqrt5=2\sqrt{10}+2\sqrt5$
Do $\Delta ABC$ vuông cân đỉnh $B\Rightarrow $ trung tuyến $BK$ trùng với đường cao $BH$
$\Rightarrow BK=BH=\sqrt5$
d) $\vec{IA}=(9;4)\Rightarrow IA=\sqrt{9^2+4^2}=\sqrt{97}$,
$\vec{IB}=(8;1)\Rightarrow IB=\sqrt{8^2+1^2}=\sqrt{65}$
$\cos\widehat{AIB}=\cos(\vec{IA},\vec{IB})=\dfrac{\vec{IA}.\vec{IB}}{IA.IB}=\dfrac{9.8+4.1}{\sqrt{97}.\sqrt{65}}=\dfrac{76}{\sqrt{6305}}$
$\Rightarrow \widehat{AIB}=16,84^o$
$\vec{AC}=(-4;-2)\Rightarrow AC=\sqrt{(-4)^2+(-2)^2}=2\sqrt5$
$\vec{AI}=(-9;-4)$, $AI=\sqrt{97}$
$\cos\widehat{IAC}=\cos(\vec{AC},\vec{AI})=\dfrac{\vec{AC}.\vec{AI}}{AC.AI}=\dfrac{-4(-9)+(-2)(-4)}{2\sqrt5.\sqrt{97}}=\dfrac{44}{2\sqrt{485}}$
$\Rightarrow \widehat{IAC}=2,6^o$
