December 17, 2018 December 17, 2018 06:00 PM PST 07:00 PM PST Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong. December 17, 2018 December 17, 2018 10:00 PM PST 11:00 PM PST From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51263

For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
29 Oct 2014, 07:25
Question Stats:
75% (01:09) correct 25% (01:28) wrong based on 385 sessions
HideShow timer Statistics
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)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 21 Jul 2014
Posts: 125

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
04 Nov 2014, 08:34
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 xcoordinate, 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 ycoordinate, 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!



Intern
Joined: 04 Jul 2014
Posts: 46

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
09 Nov 2014, 12:57
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.
Thus answer B [3,6]. Please give kudos if this helps.



Manager
Joined: 11 Nov 2011
Posts: 63
Location: United States
Concentration: Finance, Human Resources
GPA: 3.33
WE: Consulting (NonProfit and Government)

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
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

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
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 Answer is B
_________________
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 semiretired.



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]
Show Tags
15 Jul 2016, 13:10
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 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.
_________________
Send some kudos this way if I was helpful! !



Director
Joined: 04 Jun 2016
Posts: 571

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
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 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)
_________________
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 semiretired.



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]
Show Tags
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

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
21 Sep 2018, 22:30
let the coordinates 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]
Show Tags
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 xcoordinate, 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 ycoordinate, 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



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82

Re: For any triangle T in the xy–coordinate plan, the center of T is defin
[#permalink]
Show Tags
26 Sep 2018, 00:56
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 xcoordinate, 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 ycoordinate, 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 dave13the 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






