[tex]\it AB^2=(x_B-x_A)^2+(y_B-y_A)^2=(0-2)^2+(2-1)^2=4+1=5\\ \\ BC^2=(x_C-x_B)^2+(y_C-y_B)^2=(2-0)^2+(6-2)^2=4+16=20\\ \\ AC^2=(x_C-x_A)^2+(y_C-y_A)^2=(2-2)^2+(6-1)^2=25\\ \\ \\ AB^2+BC^2=5+20=25=AC^2\Rightarrow AB^2+BC^2=AC^2\ \ \ \ (*)[/tex]
[tex]\it Din\ rela\c{\it t}ia\ (*), cu\ reciproca\ teoremei\ lui\ Pitagora,\ rezult\breve{a}:\\ \\ \Delta BAC-dreptunghic,\ m(\hat{B})=90^o.\\ \\ \\ AB^2=5\Rightarrow AB=\sqrt5;\ \ BC^2=20\Rightarrow BC=\sqrt{20}=\sqrt{4\cdot5}=2\sqrt5;\\ \\ AC^2=25 \Rightarrow AC = 5.\\ \\ \\ \mathcal{P}=AB+BC+AC=\sqrt5+2\sqrt5+5=3\sqrt5+5\\ \\ \mathcal{A} = \dfrac{c_1\cdot c_2}{2}=\dfrac{AB\cdot BC}{2}=\dfrac{\sqrt5\cdot\not2\sqrt5}{\not2}=\sqrt5\cdot\sqrt5=5[/tex]
Ortocentrul triunghiului este B(0, 2), vârful unghiului drept.
Centrul cercului circumscris triunghiului este mijlocul ipotenuzei AC
[tex]\it Fie\ O(x, y),\ mijlocul\ lui\ AC.\\ \\ x=\dfrac{x_A+x_C}{2} =\dfrac{2+2}{2}=\dfrac{4}{2}=2\\ \\ \\ y=\dfrac{y_A+y_C}{2} =\dfrac{1+6}{2}=\dfrac{7}{2}=3,5\\ \\ \\ Deci,\ vom\ avea:\ \ \ O(2;\ \ 3,5)[/tex]
