The triangle

The triangle is given by three vertices:
A [0.0]
B [-4.2]
C [-6.0]
Calculate V (intersection of heights), T (center of gravity), O - center of a circle circumscribed

Correct result:

x₁ =  -3.3333
y₁ =  0.6667
x₂ =  -3
y₂ =  -1

Solution:

$$\begin{gathered} xA=0 \\ yA=0 \\ xB=-4 \\ yB=2 \\ xC=-6 \\ yC=0 \\ T(x_1,y_1) \\ {x}_1={\displaystyle \frac{xA+xB+xC}{3}}={\displaystyle \frac{0+(-4)+(-6)}{3}}=-{\displaystyle \frac{10}{3}}=-3.3333 \end{gathered}$$

$y_1={\displaystyle \frac{yA+yB+yC}{3}}={\displaystyle \frac{0+2+0}{3}}={\displaystyle \frac{2}{3}}=0.6667$

$O(x_{2},y_{2}) \ \\ \ \\ D=2 \cdot \ ( xA \cdot \ (yB - yC) + xB \cdot \ (yC - yA) + xC \cdot \ (yA - yB))=2 \cdot \ ( 0 \cdot \ (2 - 0) + (-4) \cdot \ (0 - 0) + (-6) \cdot \ (0 - 2))=24 \ \\ \ \\ x_{2}=((xA^2 + yA^2) \cdot \ (yB-yC) + (xB^2 + yB^2) \cdot \ (yC - yA) + (xC^2 + yC^2) \cdot \ (yA - yB)) / D=((0^2 + 0^2) \cdot \ (2-0) + ((-4)^2 + 2^2) \cdot \ (0 - 0) + ((-6)^2 + 0^2) \cdot \ (0 - 2)) / 24=-3$

$y_{2}=((xA^2 + yA^2) \cdot \ (xC - xB) + (xB^2 + yB^2) \cdot \ (xA - xC) + (xC^2 + yC^2) \cdot \ (xB - xA)) / D=((0^2 + 0^2) \cdot \ ((-6) - (-4)) + ((-4)^2 + 2^2) \cdot \ (0 - (-6)) + ((-6)^2 + 0^2) \cdot \ ((-4) - 0)) / 24=-1$

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