Respuesta :

frika
The point of intersection of medians is a center of triangle ABC. That means that [tex]Q(x,y)=( \frac{x_A+x_B+x_C}{3} , \frac{y_A+y_B+y_C}{3} )[/tex].
If A(1,-5), B(3,-2) and C(7,-5) the Q coordinates are:
[tex]x= \frac{1+3+7}{3} = \frac{11}{3} [/tex]
[tex]y= \frac{-5-2-5}{3} =-4[/tex]
Answer: [tex]Q( \frac{11}{3},-4 )[/tex] and the correct choice is D.
Felecityyy
The answer is the fourth choice - (3.67, -4)

Since Q is the center of gravity of the triangle ABC, we can use this relationship - vecQA + vecQB + vecQC = vec0 : we will deduct the coordinates of A, B, and C from x and y of Q to get its coordinate.

vecQA = (x-3, y+2)
vecQB = (x-1, y+5)
vecQC = (x-7, y+5)
vec0 = (0, 0)

This will be equivalent to
(x-3) + (x-1) + (x-7) = 0, and (y+2) + (y+5) + (y+5) = 0
3x - 11 = 0, and 3y + 12 = 0
x = 11/3 or 3.67, and y = 12/3 or 4

The coordinates of Q are (3.67, 4)

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