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Average of x Coordinates

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by Abdulla » Sat Nov 14, 2009 3:25 pm
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)
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by papgust » Sun Nov 15, 2009 9:38 pm
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).
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by Abdulla » Sun Nov 15, 2009 10:15 pm
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).
The OA is B, but I can't get how did you got x = 3 since x/3=3 ??
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by papgust » Sun Nov 15, 2009 10:56 pm
Its (0+6+x)/3=3
6+x = 9
x=3.

Hope this clears!
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by Abdulla » Mon Nov 16, 2009 5:37 pm
papgust wrote:Its (0+6+x)/3=3
6+x = 9
x=3.

Hope this clears!

:oops: :oops: I am sorry. I think break is needed. :lol:
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