For any point T in the xy-plane, the centre of T is defined to be the point whose x-coordinate is the average 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 triangle has vertices (0,0) and (6,0) and its centre at (3,2), what are the coordinates of the remaining vertice?
(3,4)
(3,6)
(4,9)
(6,4)
(9,6)
OA is B, shouldn't it be A????
Plane Problem
This topic has expert replies
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
The center, (3,2), is the average of the co-ordinates of the three vertices of the triangle, (0,0), (6,0), and some unknown point, let's say (c, d). So, averaging x-co-ordinates:ssraf wrote:For any point T in the xy-plane, the centre of T is defined to be the point whose x-coordinate is the average 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 triangle has vertices (0,0) and (6,0) and its centre at (3,2), what are the coordinates of the remaining vertice?
(3,4)
(3,6)
(4,9)
(6,4)
(9,6)
OA is B, shouldn't it be A????
3 = (0+6+c)/3
9 = 6 + c
3 = c
and averaging y-co-ordinates
2 = (0+0+d)/3
6 = d
So the remaining vertex is (3, 6).