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 answer:

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

Step-by-step explanation:

$$\begin{gathered} A=(0,0) \\ B=(-4,2) \\ C=(-6,0) \\ T=(x_1,y_1) \\ {x}_1={\displaystyle \frac{A_x+B_x+C_x}{3}}={\displaystyle \frac{0+(-4)+(-6)}{3}}=-{\displaystyle \frac{10}{3}}=-3\frac{1}{3}=-3.3333 \end{gathered}$$

$y_1={\displaystyle \frac{A_y+B_y+C_y}{3}}={\displaystyle \frac{0+2+0}{3}}={\displaystyle \frac{2}{3}}=0.6667$

$$\begin{gathered} O(x_2,y_2) \\ D=2\cdot (A_x\cdot (B_y-C_y)+B_x\cdot (C_y-A_y)+C_x\cdot (A_y-B_y))=2\cdot (0\cdot (2-0)+(-4)\cdot (0-0)+(-6)\cdot (0-2))=24 \\ {x}_2=((A_x^2+A_y^2)\cdot (B_y-C_y)+(B_x^2+B_y^2)\cdot (C_y-A_y)+(C_x^2+C_y^2)\cdot (A_y-B_y))\mathrm{/}D=((0^2+0^2)\cdot (2-0)+((-4{)}^2+2^2)\cdot (0-0)+((-6{)}^2+0^2)\cdot (0-2))\mathrm{/}24=-3 \end{gathered}$$

$y_2=((A_x^2+A_y^2)\cdot (C_x-B_x)+(B_x^2+B_y^2)\cdot (A_x-C_x)+(C_x^2+C_y^2)\cdot (B_x-A_x))\mathrm{/}D=((0^2+0^2)\cdot ((-6)-(-4))+((-4{)}^2+2^2)\cdot (0-(-6))+((-6{)}^2+0^2)\cdot ((-4)-0))\mathrm{/}24=-1$

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