It is currently 18 Nov 2017, 03:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M15-19

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132554 [0], given: 12326

### Show Tags

16 Sep 2014, 00:55
Expert's post
12
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

69% (02:26) correct 31% (02:36) wrong based on 77 sessions

### HideShow timer Statistics

If on the coordinate plane $$(6, 2)$$ and $$(0, 6)$$ are the endpoints of the diagonal of a square, what is the distance between point $$(0, 0)$$ and the closest vertex of the square?

A. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 132554 [0], given: 12326

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132554 [3], given: 12326

### Show Tags

16 Sep 2014, 00:55
3
KUDOS
Expert's post
4
This post was
BOOKMARKED
Official Solution:

If on the coordinate plane $$(6, 2)$$ and $$(0, 6)$$ are the endpoints of the diagonal of a square, what is the distance between point $$(0, 0)$$ and the closest vertex of the square?

A. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

Given endpoints of diagonal of a square: $$B(0,6)$$ and $$D(6,2)$$. Let other vertices be $$A$$ (closest to the origin) and $$C$$ (farthest to the origin):

Length of the diagonal would be: $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}$$

Coordinates of the midpoint $$M$$ of the diagonal would be: $$M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)$$.

{The slope of line segment AM}*{The slope of line segment BD}=-1 (as they are perpendicular to each other), so $$\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1$$, which simplifies to $$y-4=\frac{3}{2}(x-3)$$

The distance between the unknown vertices to the midpoint is half the diagonal:

$$(x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13$$;

$$(x-3)^2+\frac{9}{4}(x-3)^2=13$$;

$$(x-3)^2=4$$;

Which gives: $$x=1$$ and $$y=1$$ OR $$x=5$$ and $$y=7$$.

Hence point $$A$$ is at $$(1,1)$$ and point $$C$$ is at $$(5,7)$$. Closest to the origin is $$A$$. Distance $$OA=\sqrt{2}$$

_________________

Kudos [?]: 132554 [3], given: 12326

Manager
Joined: 11 Aug 2011
Posts: 194

Kudos [?]: 400 [0], given: 886

Location: United States
Concentration: Economics, Finance
GMAT Date: 10-16-2013
GPA: 3
WE: Analyst (Computer Software)

### Show Tags

29 Aug 2015, 01:50
Hi

Is there any shorter way to solve this problem.
I have understood the solution but this seems to be a very tough and complex problem.
I am wondering if this can be solved in some other way.
_________________

Kudos me if you like my post !!!!

Kudos [?]: 400 [0], given: 886

Intern
Joined: 20 Sep 2014
Posts: 6

Kudos [?]: 3 [0], given: 1

### Show Tags

13 Nov 2015, 11:54
2
This post was
BOOKMARKED
Hi akhil911, maybe this will help. I calculated the midpoint in the same way as described by Bunuel to find (3,4). Then I determined the delta from point B to the midpoint M as (0,6) - (3,4) = (3,2). In order to find point A simply swap the x and y and subtract from the midpoint: (3,4) - (2,3) = (1,1). Which gives √2 as distance.

Kudos [?]: 3 [0], given: 1

Intern
Joined: 04 Aug 2014
Posts: 5

Kudos [?]: 2 [1], given: 2

### Show Tags

26 Dec 2015, 07:19
1
KUDOS
1
This post was
BOOKMARKED
Or for points (6,2) and (0,6) in this case:
Absolute difference for x is 6-0=6
Absolute difference for y is 6-2=4
Using logic, since this is a square, sides need to be equal and average of 6+4 is 5
Hence, absolute difference of 5 units away is for x is x=1 and for y is y=1, with intersection at point (1, 1) being equidistant to B and D
Then work out the distance as before

Kudos [?]: 2 [1], given: 2

Manager
Joined: 18 May 2016
Posts: 67

Kudos [?]: 63 [0], given: 105

GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)

### Show Tags

30 May 2016, 05:12
Shorter solution if one can draw a correct sketch of the figure:

The shortest distance from (0,0) to one of the vertexes of the square is perpendicural distance to A. A and (0,0) forms a square with s=1. The shortest distance will be the diagonal of that square which is V2s or 1 x V2 = V2 the answer.

Even if the graph is drawn somewhat right, we are looking for the answer that has nV2 in it.
_________________

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!

Kudos [?]: 63 [0], given: 105

Manager
Joined: 25 Sep 2015
Posts: 145

Kudos [?]: 12 [0], given: 74

Location: United States
GMAT 1: 700 Q48 V37
GPA: 3.26

### Show Tags

06 Jul 2016, 14:16
The distance between the unknown vertices to the midpoint is half the diagonal:

(x−3)2+(y−4)2=(52√2)2=13(x−3)2+(y−4)2=(522)2=13;

I don't understand what this step is doing?

Kudos [?]: 12 [0], given: 74

Intern
Joined: 19 Mar 2013
Posts: 3

Kudos [?]: [0], given: 3

### Show Tags

27 Jul 2016, 10:59
rachitshah
It seems like the steps from the given explanation accomplish the following:
1) Find the distance of the original diagonal: (6-0)^2 + (2-6)^2 = X^2; X = root 52
2) Find the midpoint of the original diagonal: (6+0)/2, (6+2)/2 = (3,4)
3) Find the equation for the line for the other diagonal, since this will be the line that has the coordinates closest to (0,0)
Key to this step: This line is the perpendicular bisector of the diagonal (the diagonals of squares are perpendicular bisectors = fact in geometry)
You can find the equation for the line the other diagonal by doing the following:
a) y = mx + b (generic equation for line)
b) find the slope of the original diagonal: (6-2)/(0-6) = -2/3 and take the negative reciprocal of this = 3/2 since the other diagonal is perpendicular
c) now you have y = 3x/2 + b; find the value of b by plugging in (3,4) and you get y = 3x/2 - 1/2, which is the same thing as y-4 = 3/2(x-3)
4) Distance from farthest coordinates to midpoint of diagonal = Distance from shortest coordinates to midpoint of diagonal = 1/2 the length of the diagonal (in this case, (root52/2)^2).
5) The distance from the farthest or shortest coordinates of the diagonal to the midpoint can be solved by using pythagorean theorem:
(x-3)^2 +(y-4)^2 = (root52/2)^2. (root52/2)^2 = 13
6) From step 3 you know that y-4 = 3/2(x-3). Plug this into y for step 5 above ==> (x-3)^2 + (y-4)^2 = 13 --> (x-3)^2 + [3/2(x-3)]^2 = 13 -->
(x-3)^2 +9/4(x-3)^2 = 13; solving for x gives you (x-3)^2 = 4, so x can be 1 or 5.

Kudos [?]: [0], given: 3

Senior Manager
Joined: 31 Mar 2016
Posts: 406

Kudos [?]: 81 [0], given: 197

Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

28 Aug 2016, 09:47
I think this is a high-quality question and I agree with explanation.

Kudos [?]: 81 [0], given: 197

Intern
Joined: 11 Nov 2014
Posts: 40

Kudos [?]: 16 [0], given: 102

Concentration: Marketing, Finance
WE: Programming (Computer Software)

### Show Tags

24 Oct 2016, 00:39
rachitshah wrote:
The distance between the unknown vertices to the midpoint is half the diagonal:

(x−3)2+(y−4)2=(52√2)2=13(x−3)2+(y−4)2=(522)2=13;

I don't understand what this step is doing?

This step is finding the distance from vertex A(x,y) to mid point M(3,4) or the length of AM you can say.
since AC is also a diagonal, AM will be half its length (which is\sqrt{52} and diagonals of a square are equal in length.)

I hope it clears your doubt.

Kudos [?]: 16 [0], given: 102

Manager
Joined: 27 Jan 2016
Posts: 150

Kudos [?]: 67 [0], given: 123

Schools: ISB '18

### Show Tags

10 Aug 2017, 23:29
Skytail wrote:
Hi akhil911, maybe this will help. I calculated the midpoint in the same way as described by Bunuel to find (3,4). Then I determined the delta from point B to the midpoint M as (0,6) - (3,4) = (3,2). In order to find point A simply swap the x and y and subtract from the midpoint: (3,4) - (2,3) = (1,1). Which gives √2 as distance.

Can you pls explain the logic behind this approach. What do you mean by delta here.

Kudos [?]: 67 [0], given: 123

Manager
Joined: 27 Jan 2016
Posts: 150

Kudos [?]: 67 [0], given: 123

Schools: ISB '18

### Show Tags

10 Aug 2017, 23:41
Bunuel wrote:
Official Solution:

If on the coordinate plane $$(6, 2)$$ and $$(0, 6)$$ are the endpoints of the diagonal of a square, what is the distance between point $$(0, 0)$$ and the closest vertex of the square?

A. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

Given endpoints of diagonal of a square: $$B(0,6)$$ and $$D(6,2)$$. Let other vertices be $$A$$ (closest to the origin) and $$C$$ (farthest to the origin):

Length of the diagonal would be: $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}$$

Coordinates of the midpoint $$M$$ of the diagonal would be: $$M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)$$.

{The slope of line segment AM}*{The slope of line segment BD}=-1 (as they are perpendicular to each other), so $$\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1$$, which simplifies to $$y-4=\frac{3}{2}(x-3)$$

The distance between the unknown vertices to the midpoint is half the diagonal:

$$(x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13$$;

$$(x-3)^2+\frac{9}{4}(x-3)^2=13$$;

$$(x-3)^2=4$$;

Which gives: $$x=1$$ and $$y=1$$ OR $$x=5$$ and $$y=7$$.

Hence point $$A$$ is at $$(1,1)$$ and point $$C$$ is at $$(5,7)$$. Closest to the origin is $$A$$. Distance $$OA=\sqrt{2}$$

Hi Bunuel,

Is there any shorter way to solve this problem. Also, can you pls explain the logic behind the Skytail's approach in one of the above posts.

Kudos [?]: 67 [0], given: 123

Manager
Joined: 12 Sep 2016
Posts: 71

Kudos [?]: 9 [0], given: 796

Location: India
Schools: Kellogg '20
GMAT 1: 700 Q50 V34
GPA: 3.33

### Show Tags

29 Aug 2017, 07:24
this is a very very lengthy problem!
It's great for practice though!

Kudos [?]: 9 [0], given: 796

Intern
Joined: 18 Dec 2015
Posts: 1

Kudos [?]: [0], given: 0

### Show Tags

09 Nov 2017, 16:57
Quicker way to solve this:

The midpoint of the diagonal is (3,4). The vector from the right bottom corner to the midpoint is V1 = <-3, -2>. You need to find a vector perpendicular to this vector and of the same magnitude and add it to the midpoint position vector to find the corners.

Vectors perpendicular meet:

-3x - 2y = 0 or y = (3/2) x
So the form is <x, 3/2x>

The two perpendicular vectors that have the same magnitude as V1 will meet:

x^2 + [(3/2)x]^2 = 13
x = +/- 2

So the two vectors are <2,3> or <-2,-3>

To find the corner closest to the origin add <-2,-3> to the position vector of the midpoint <3,4> to get <1,1>.

The magnitude of this vector sqrt(2) is the answer.

Kudos [?]: [0], given: 0

M15-19   [#permalink] 09 Nov 2017, 16:57
Display posts from previous: Sort by

# M15-19

Moderators: Bunuel, chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.