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For any triangle T in the xy-coordinate plane, the center of

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Manager
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Kudos [?]: 42 [0], given: 13

For any triangle T in the xy-coordinate plane, the center of [#permalink] New post 11 Sep 2011, 11:06
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A
B
C
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  45% (medium)

Question Stats:

67% (01:54) correct 33% (01:04) wrong based on 12 sessions
For any triangle T in the xy-coordinate plane, the center of T is defined to be the point whose x-coord.is the avr. of the x coord-s of the vertices T and whose y-coord. is the avr. of the y-coord-s of the vert.T. If a certain triangle has vertices at the points (0;0) and (6;0) and the center at the point (3;2), what are the coord-s of the remain.vertex"

(3;4)
(3;6)
(4;9)
(6;4)
(9;6)
[Reveal] Spoiler: OA
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Senior Manager
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Kudos [?]: 120 [1] , given: 22

Re: triangle on the xy coord.plane [#permalink] New post 11 Sep 2011, 13:01
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Three vertices are A(0,0) B(6,0) and C(x,y). We need to find x,y
Center of triangle = O(3,2)
Now co-ordinate of center of triangle is an average of co-ordinate of vertices.
X-co-ordinate of O is = X-co-ordinate of A + X-co-ordinate of B + X-co-ordinate of C / 3
3= 0+6+x/3
9=6+x
x=3

Similarly, Y-co-ordinate of O is = Y-co-ordinate of A + Y-co-ordinate of B + Y-co-ordinate of C / 3
2=0+0+y/3
y=6

Hence (x,y) = (3,6)

OA B.
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Kudos [?]: 58 [0], given: 42

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Re: triangle on the xy coord.plane [#permalink] New post 11 Sep 2011, 13:38
by definition center = ( ((x1+x2+x)/3) , ((y1+y2+y3)/3) )

two given vertices are (0,0) and (6,0). let (x,y) be the unknown third vertex.

Also given center = (3,2)

=> (3,2)= ( (6+x)/3 , y/3 )

equation LHS with RHS , we have

y/3 = 2 => y=6

6+x = 9 => x=3

=> third vertex = (3,6)

Answer is B.
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Kudos [?]: 16 [0], given: 21

Re: triangle on the xy coord.plane [#permalink] New post 12 Sep 2011, 01:43
Easy one. It's B. You would have used wrong co-ordinates...
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Re: triangle on the xy coord.plane   [#permalink] 12 Sep 2011, 01:43
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For any triangle T in the xy-coordinate plane, the center of

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