Question

2-) Determine o Baricentro dos Triângulos de Vértices :
a-) A( 2, 5) ,B( 4, 6) e C ( 9, 4)
b-) A( 0, 7) , B ( 2, -5 ) e C (13, 13)

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Answer 1

a. G (5, 5)

$gx = \frac{2 + 4 + 9}{3} = \frac{15}{3} = 5$

$gy = \frac{5 + 6 + 4}{3} = \frac{15}{3} = 5$

b. G (5, 5)

$gx = \frac{0 + 2 + 13}{3} = \frac{15}{3} = 5$

$gy = \frac{7 - 5 + 13}{3} = \frac{15}{ 3 } = 5$

Answer 2

$\Large\boxed{\begin{array}{l}\sf Seja~A(x_A,y_A),B(x_B,y_B)~e~C(x_C,y_C)\\\sf as~coordenadas~de~um~tri\hat angulo\\\sf qualquer~no~plano~cartesiano.\\\sf O~baricentro~G(x_G,y_G)~deste~tri\hat angulo~\acute e~dado~por\\\sf x_G=\dfrac{x_A+x_B+x_C}{3}~e~y_G=\dfrac{y_A+y_B+y_C}{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt a)~\sf A(2,5),B(4,6)~e~C(9,4)\\\sf x_G=\dfrac{2+4+9}{3}=\dfrac{15}{3}=5\\\\\sf y_G=\dfrac{5+6+4}{3}=\dfrac{15}{3}=5\\\sf G(5,5)\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf A(0,7),B(2,-5)~e~C(13,13)\\\sf x_G=\dfrac{0+2+13}{3}=\dfrac{15}{3}=5\\\\\sf y_G=\dfrac{7-5+13}{3}=\dfrac{15}{3}=5\\\sf G(5,5)\end{array}}$