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:

xA=0 yA=0 xB=4 yB=2 xC=6 yC=0  T(x1,y1)  x1=xA+xB+xC3=0+(4)+(6)3=103=3.3333
y1=yA+yB+yC3=0+2+03=23=0.6667
O(x2,y2)  D=2 (xA (yByC)+xB (yCyA)+xC (yAyB))=2 (0 (20)+(4) (00)+(6) (02))=24  x2=((xA2+yA2) (yByC)+(xB2+yB2) (yCyA)+(xC2+yC2) (yAyB))/D=((02+02) (20)+((4)2+22) (00)+((6)2+02) (02))/24=3O(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
y2=((xA2+yA2) (xCxB)+(xB2+yB2) (xAxC)+(xC2+yC2) (xBxA))/D=((02+02) ((6)(4))+((4)2+22) (0(6))+((6)2+02) ((4)0))/24=1y_{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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