Question

5. Se (2,1), (3,3) e (6,2) são os pontos médios dos lados de um triangulo, quais são os seus
vértices?
a) (-1,2),(5,0),(7,4) b) (2,2), (2,0), (4,4) c) (1,1), (3,1), (5,5) d) (3,1), (1,1),
(3,5)

Answer 1

(2,1), (3,3) e (6,2)

A(xa,ya)
B(xb,yb)
C(xc,yc)

(2,1) ponto medio de A e B

(xa + xb , ya + yb) = 2(2,1)
(xa + xb , ya + yb) = (4,2)

xa + xb = 4
ya + yb = 2

(3,3) ponto medio de B e C

(xb + xc , yb + yc) = 2(3,3)
(xb + xc , yb + yc) = (6,6)

xb + xc = 6
yb + yc = 6

(6,2) ponto medio de A e C

(xa + xc , ya + yc) = 2(6,2)
(xa + xc , ya + yc) = (12,4)

xa + xc = 12
ya + yc = 4

temos 6 equacoes:

xa + xb = 4 ①
ya + yb = 2 ②
xb + xc = 6 ③
yb + yc = 6 ④
xa + xc = 12 ⑤
ya + yc = 4 ⑥


somando as equacoes ① com ⑤
xa + xb = 4
xa + xc = 12

xa + xa + xb + xc = 16
2xa + xb + xc = 16

sendo xb + xc = 6

2xa + 6 = 16
2xa = 16 - 6
xa = 10/2
xa = 5

equacao ①
xa + xb = 4
5 + xb = 4
xb = 4 - 5
xb = - 1

equacao ③
xb + xc = 6
- 1 + xc = 6
xc = 6 + 1
xc = 7

somando as equacoes ② com ④
ya + yb = 2
yb + yc = 6

ya + yb + yb + yc = 8
2yb + ya + yc = 8

sendo ya + yc = 4

2yb + 4 = 8
2yb = 8 - 4
yb = 4/2
yb = 2

equacao ②

ya + yb = 2
ya + 2 = 2
ya = 2 - 2
ya = 0

equacao ④

yb + yc = 6
2 + yc = 6
yc = 6 - 2
yc = 4

portanto os pontos sao:

A(5,0)
B(- 1,2)
C(7,4)

R.: Letra A