The triangle

Three vertices give the triangle:
A [0.0]
B [-4.2]
C [-6.0]
Calculate V (intersection of heights), T (center of gravity), O - the center of a circle circumscribed

Correct answer:

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

Step-by-step explanation:

A = (0, 0) B = (- 4, 2) C = (- 6, 0) T = (x_{1}, y_{1}) x_{1} = \frac{A _{x} + B _{x} + C _{x}}{3} = \frac{0 + ( - 4 ) + ( - 6 )}{3} = - \frac{1 0}{3} = - 3 \frac{1}{3} = - 3.3333

y_{1} = \frac{A _{y} + B _{y} + C _{y}}{3} = \frac{0 + 2 + 0}{3} = \frac{2}{3} = 0.6667

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})) / 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} = ((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})) / 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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