Abdulla wrote:For any triangle T in the xy-coordinate plane, the center of T is defined to be the point whose x-coordinate is the average (arithmetic mean) of the x-coordinates of the vertices of T and whose y-coordinate is the average of the y-coordinates of the vertices of T.If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A.(3,4)
B.(3,6)
C.(4,9)
D.(6,4)
E.(9,6)

papgust wrote:

Abdulla wrote:For any triangle T in the xy-coordinate plane, the center of T is defined to be the point whose x-coordinate is the average (arithmetic mean) of the x-coordinates of the vertices of T and whose y-coordinate is the average of the y-coordinates of the vertices of T.If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A.(3,4)
B.(3,6)
C.(4,9)
D.(6,4)
E.(9,6)

Answer is B.
The questions basically says that centroid of triangle is at the point (3,2). You are given 2 vertices in the triangle. You are asked to find the third vertex.

Apply this formula of finding centroid of a triangle,
[(x1 + x2 + x3)/3, (y1 + y2 + y3)/3] = (3,2)
[(0+6+x)/3, (0+0+y)/3] = (3,2)

Equate individually,
(0+6+x)/3 = 3
x=3

(0+0+y)/3 = 2
y=6

Hence (3,6).

papgust wrote:Its (0+6+x)/3=3
6+x = 9
x=3.

Hope this clears!