a. $\vec{AC}=(3,1)$
$\Rightarrow$ Đường thẳng AC nhận vtcp $\vec{AC}=(3,1)$
Ta có (AC): - qua $A(-1,3)$
- vtcp $\vec{AC}=(3,1)$
$\Rightarrow$ ptts: $\left\{ \begin{array}{l}x=-1+3t\\y=3+t\end{array} \right.$
b. Ta có: $BH \perp AC$
$\Rightarrow$ BH nhận vtpt là $\vec{AC}=(3,1)$
Ta có (BH): - qua $B(0,2)$
- vtpt $\vec{AC}=(3,1)$
$\Rightarrow$ pttq: $3x+y-2=0$
c. $\vec{AC}=(3,1)$
$\Rightarrow AC=\sqrt{3^2+1^2}=\sqrt{10}$
Ta có (C): - tâm $A(-1,3)$
- $R=AC=\sqrt{10}$
$\Rightarrow$ ptdtron: $(x+1)^2+(y-3)^2=10$