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# For any triangle T in the xy–coordinate plan, the center of T is defin

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Math Expert
Joined: 02 Sep 2009
Posts: 51263
For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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29 Oct 2014, 07:25
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6
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Difficulty:

25% (medium)

Question Stats:

75% (01:09) correct 25% (01:28) wrong based on 385 sessions

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Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, 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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Joined: 21 Jul 2014
Posts: 125
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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04 Nov 2014, 08:34
3
1
Hi Bunuel,

Great question. The best mode of attack is to Dive In.

Let's start by labeling the three vertices of triangle T as a, b, and c. We know pt a is at (0,0), b is at (6,0), and c is unknown, (cx, cy).

The center, as defined in the problem, is the arithmetic mean of the x and y coordinates individually. Let's write that out as a formula, where the center of triangle T is labeled as pt m at (mx, my).

mx = (ax + bx + cx)/3
my = (ay + by + cy)/3

We know from the problem that pt m is at (3,2), so let's plug in first for the x-coordinate, cx:
3 = (0 + 6 + cx)/3
With some arithmetic, we can solve this to see that cx = 3.

Now, let's do the same for the y-coordinate, cy:
2 = (0 + 0 + cy)/3
So... cy = 6.

Putting these together, we now have that the coordinate of the missing vertex is at (cx, cy), or (3,6)... Answer Choice B.

Hope this helps!
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Joined: 04 Jul 2014
Posts: 46
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Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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09 Nov 2014, 12:57
1
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, 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)

Notice that (3,2) lies on one the median which divides the side connecting [0,0] & [6,0]....and it is give [3,2] is the centre...we know that centriod will divide the median in the ratio 2:1 .....the smaller part of the ratio is 2 here, so the bigger must be 4....and hence the point is [3,6]. Now add the x of all vertices and y of all vertices , find the averages and cross check.

------------
Please give kudos if this helps.
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Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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27 Apr 2015, 06:42
Can I solve this way?

let the vertices be A, B, and C
A=(0,0), B=(6,0), and since the central is (3,2) then C=(3,n)

sum of x=0+6+3=9; avg=9/3=3
sum of y=0+0+n=; avg=n/3=2 (since the center is (3,2))
thus n=3*2=6

So the answer is (3,6) B
Director
Joined: 04 Jun 2016
Posts: 571
GMAT 1: 750 Q49 V43
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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14 Jul 2016, 10:12
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, 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)

Seems like a difficult question , but actually it can be solved in less than 30 seconds.
First plot all the given (X,Y) pair.
You will get the base of the triangle with length of 6 and mid point of triangle at 3,3
Now the vortex will be double at the Y coordinate of the midpoint
Double of 3= 6 (y=3)
Average of X is already 3
Therefore the X,Y pair will become 3,6

_________________

Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

Intern
Joined: 26 Aug 2015
Posts: 34
Concentration: Strategy, Economics
GMAT 1: 570 Q40 V28
GMAT 2: 740 Q49 V41
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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15 Jul 2016, 13:10
1
LogicGuru1 wrote:
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, 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)

Seems like a difficult question , but actually it can be solved in less than 30 seconds.
First plot all the given (X,Y) pair.
You will get the base of the triangle with length of 6 and mid point of triangle at 3,3
Now the vortex will be double at the Y coordinate of the midpoint
Double of 3= 6 (y=3)
Average of X is already 3
Therefore the X,Y pair will become 3,6

The center is 3,2.

Actually the average is of 3 data points, that is why it is 6 because (0+0+y)/3=2 y=6.

X remains at 3 because
(0 + 6 + x) /3 = 3
6+x=9
x=3

At least it made sense to me this way.
_________________

Send some kudos this way if I was helpful! !

Director
Joined: 04 Jun 2016
Posts: 571
GMAT 1: 750 Q49 V43
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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15 Jul 2016, 20:58
Ilomelin wrote:
LogicGuru1 wrote:
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

For any triangle T in the xy–coordinate plan, 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)

Seems like a difficult question , but actually it can be solved in less than 30 seconds.
First plot all the given (X,Y) pair.
You will get the base of the triangle with length of 6 and mid point of triangle at 3,3
Now the vortex will be double at the Y coordinate of the midpoint
Double of 3= 6 (y=3)
Average of X is already 3
Therefore the X,Y pair will become 3,6

The center is 3,2.

Actually the average is of 3 data points, that is why it is 6 because (0+0+y)/3=2 y=6.

X remains at 3 because
(0 + 6 + x) /3 = 3
6+x=9
x=3

At least it made sense to me this way.

Yup , you are right

Average should should be 0+3+6=9/3
= 3
Double of 3=6
so vortex will be (x,y)=(3,6)
_________________

Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

Manager
Joined: 24 Dec 2016
Posts: 93
Location: India
Concentration: Finance, General Management
WE: Information Technology (Computer Software)
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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22 Feb 2017, 17:06
I think its a very simple averages question. Median point (3,2) should be equal to {(x1+x2+x3)/3, (y1+y2+y3)/3}.
We have (x1,y1) and (x2,y2) as (0,0) and (6,0).
Thus, 3= (0+6+x3)/3 => x3 = 3
& 2= (0+0+y3)/3 => y3 = 6.
The point, therefore, is (3,6). Option B.
Manager
Joined: 20 Jun 2017
Posts: 93
GMAT 1: 570 Q49 V19
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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21 Sep 2018, 22:30
let the co-ordinates of the 3rd vertex be (x,y)

3 = $$\frac{(0+6+x)}{3}$$

x = 3

2 = $$\frac{(0+0+y)}{3}$$

y = 6

Hence the 3rd vertex is (3,6)
VP
Joined: 09 Mar 2016
Posts: 1247
Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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23 Sep 2018, 03:26
LighthousePrep wrote:
Hi Bunuel,

Great question. The best mode of attack is to Dive In.

Let's start by labeling the three vertices of triangle T as a, b, and c. We know pt a is at (0,0), b is at (6,0), and c is unknown, (cx, cy).

The center, as defined in the problem, is the arithmetic mean of the x and y coordinates individually. Let's write that out as a formula, where the center of triangle T is labeled as pt m at (mx, my).

mx = (ax + bx + cx)/3
my = (ay + by + cy)/3

We know from the problem that pt m is at (3,2), so let's plug in first for the x-coordinate, cx:
3 = (0 + 6 + cx)/3
With some arithmetic, we can solve this to see that cx = 3.

Now, let's do the same for the y-coordinate, cy:
2 = (0 + 0 + cy)/3
So... cy = 6.

Putting these together, we now have that the coordinate of the missing vertex is at (cx, cy), or (3,6)... Answer Choice B.

Hope this helps!

niks18 can you pls explain based on what rule do we plug in coordinates of centroid into equation to find third vertex ? i cant undestand logic. We need the coordinates of the last vertex, but we are plugging in corrdinates of centroid thank you
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Location: India
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Re: For any triangle T in the xy–coordinate plan, the center of T is defin  [#permalink]

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26 Sep 2018, 00:56
1
dave13 wrote:
LighthousePrep wrote:
Hi Bunuel,

Great question. The best mode of attack is to Dive In.

Let's start by labeling the three vertices of triangle T as a, b, and c. We know pt a is at (0,0), b is at (6,0), and c is unknown, (cx, cy).

The center, as defined in the problem, is the arithmetic mean of the x and y coordinates individually. Let's write that out as a formula, where the center of triangle T is labeled as pt m at (mx, my).

mx = (ax + bx + cx)/3
my = (ay + by + cy)/3

We know from the problem that pt m is at (3,2), so let's plug in first for the x-coordinate, cx:
3 = (0 + 6 + cx)/3
With some arithmetic, we can solve this to see that cx = 3.

Now, let's do the same for the y-coordinate, cy:
2 = (0 + 0 + cy)/3
So... cy = 6.

Putting these together, we now have that the coordinate of the missing vertex is at (cx, cy), or (3,6)... Answer Choice B.

Hope this helps!

niks18 can you pls explain based on what rule do we plug in coordinates of centroid into equation to find third vertex ? i cant undestand logic. We need the coordinates of the last vertex, but we are plugging in corrdinates of centroid thank you

Hi dave13
the question says that the x & y coordinates of the center of triangle is equal to the averages of x & y coordinates of three vertices.
Now you know the average of x coordinate and you know two x coordinates of the vertices, you need to find out the 3rd x coordinate
Similarly for y coordinate.
Take this question as an average question and try to solve
Re: For any triangle T in the xy–coordinate plan, the center of T is defin &nbs [#permalink] 26 Sep 2018, 00:56
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