For any triangle T in the xy–coordinate plan, the center of T is defin : Problem Solving (PS)

evdo

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27 Apr 2015, 07:42

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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

LogicGuru1

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14 Jul 2016, 11:12

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Bunuel

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
Answer is B

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15 Jul 2016, 14:10

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LogicGuru1

Bunuel

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
Answer is B

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.

LogicGuru1

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15 Jul 2016, 21:58

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Ilomelin

LogicGuru1

Bunuel

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
Answer is B

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)

Shruti0805

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22 Feb 2017, 18:06

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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.

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21 Sep 2018, 23:30

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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)

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23 Sep 2018, 04:26

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LighthousePrep

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

niks18

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26 Sep 2018, 01:56

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dave13

LighthousePrep

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

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01 Aug 2019, 05:39

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galiya

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)

3 = (0 + 6 + x)/3
x = 9 - 6 = 3

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

Therefore, 3rd Vertex = (3, 6) => B

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17 Sep 2021, 02:59

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I got it right and thought this was such an easy question but I am confused after seeing the replies.

My method:

X coordinate of 3rd vertex:

3= (0+6+x)/3......... because average.
x=3.

Y coordinate of 3rd vertex:

2= (0+0+y)/3......... because average.
y=6.

Thus, 3rd vertex is (3,6)

Is this a valid method or I was just lucky? Bunuel

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03 Dec 2023, 09:51

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Nothing but a word problem: need to realize the wording and basics of arithmetic mean.

Posted from my mobile device

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Would it be too risky to just draw the triangle down on an x-y coordinate plan and deduce the coordinates of the third missing vertex by just trying every solution that's given and seeing in what case point (3 ; 2) is actually the center? By doing this I was able to get the answer right pretty quickly, so I wonder if this could be a solution for geometry problems just like this, where getting into long calculations could waste you a lot of time.

Krunaal

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12 Jan 2025, 10:29

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That's another method to do it. You can absolutely do it that way if you find visualizing much easier and quicker for you.

fracur45

Would it be too risky to just draw the triangle down on an x-y coordinate plan and deduce the coordinates of the third missing vertex by just trying every solution that's given and seeing in what case point (3 ; 2) is actually the center? By doing this I was able to get the answer right pretty quickly, so I wonder if this could be a solution for geometry problems just like this, where getting into long calculations could waste you a lot of time.

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