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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

11 Jun 2013, 06:56

1

This post received KUDOS

9

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

63% (02:17) correct 37% (02:06) wrong based on 275 sessions

HideShow timer Statistics

In the rectangular coordinate system Point O has coordinates (0,0) and Point B has coordinates (4,4) and if point A is equidistant from points O and B and the area of the triangle OAB is 16, which of the following are the possible coordinates of point A.

In the rectangular coordinate system Point O has coordinates (0,0) and Point B has coordinates (4,4) and if point A is equidistant from points O and B and the area of the triangle OAB is 16, which of the following are the possible coordinates of point A.

A. (-2,6) B. (0,4) C. (2,-6) D. (2,6) E. (4,0)

Look at the diagram below:

Attachment:

Area.png [ 11.71 KiB | Viewed 4087 times ]

Since A is equidistant from points O and B, then it must be somewhere on the green line (perpendicular bisector of OB).

(2,-6) and (2,6) are not on that line. If A is at (0,4) or (4,0), then the area is 1/2*4*4=8. Thus A must be at (-2,6).

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

11 Jun 2013, 07:23

Bunuel wrote:

Since A is equidistant from points O and B, then it must be somewhere on the green line (perpendicular bisector of OB).

Thanks Bunuel for the quick response. However, I'm not able to understand the underlying logic in the above statement. I mean, why it is necessary that point A must lie on perpendicular of OB. Can you please elaborate. I'm pretty weak in coordinate geometry.

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

11 Jun 2013, 08:52

1

This post received KUDOS

1

This post was BOOKMARKED

imhimanshu wrote:

In the rectangular coordinate system Point O has coordinates (0,0) and Point B has coordinates (4,4) and if point A is equidistant from points O and B and the area of the triangle OAB is 16, which of the following are the possible coordinates of point A.

A. (-2,6) B. (0,4) C. (2,-6) D. (2,6) E. (4,0)

Assuming the base of the triangle as OB, which is of length \(4\sqrt{2}\), and let the height from A to OB be h-->

\(\frac{1}{2}*4\sqrt{2}*h\) = 16 --> h = \(4\sqrt{2}\). Also, as the point A is equidistant from both O and B, the point A will lie on the perpendicular bisector of the triangle OAB. Thus, h = the distance between the co-ordinates of A and (2,2)[the mid point of the line segment OB]. Only A satisfies for h = \(4\sqrt{2}\) A.
_________________

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

11 Jun 2013, 15:20

1

This post received KUDOS

imhimanshu wrote:

why it is necessary that point A must lie on perpendicular of OB

Hey,

Are you familiar with the term 'equidistant', meaning that the distance between OA and AB must be the same?

Imagine that A is some point that lies on the line OB, and A is equidistant from both points. Then A would have to be the midpoint of OB.

Now, extending that idea to find all points that are equidistant from O and B, if you try to plot a few points, you will see that they form a line that intersects the midpoint of OB, and extends to infinite, perpendicular to OB.

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

02 Aug 2013, 09:18

1

This post received KUDOS

another way the question can be solved is by using the distance forumula Since, \(AO=AB\), their distance will be the same, therefore by using the formula we get \(x^2+y^2=(x-4)^2+(y-4)^2\) Solving the above equation, we get \(x+y=4\). Now the only option that satisfies this equation is A.

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

02 Aug 2013, 09:24

Bunuel wrote:

imhimanshu wrote:

In the rectangular coordinate system Point O has coordinates (0,0) and Point B has coordinates (4,4) and if point A is equidistant from points O and B and the area of the triangle OAB is 16, which of the following are the possible coordinates of point A.

A. (-2,6) B. (0,4) C. (2,-6) D. (2,6) E. (4,0)

Look at the diagram below:

Attachment:

Area.png

Since A is equidistant from points O and B, then it must be somewhere on the green line (perpendicular bisector of OB).

(2,-6) and (2,6) are not on that line. If A is at (0,4) or (4,0), then the area is 1/2*4*4=8. Thus A must be at (-2,6).

Answer: A.

Hope it's clear.

I was wondering, since we wouldn't be provided with a graph paper and can only draw a rough figure, how would one exactly come to know the middle point of the line OB?

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

15 Oct 2016, 20:16

Another way of solving this.

When we draw the given senerio we realise for A to be equidistant to both o and B it needs to be on the 2nd or 4th quadrant. only option in the second quadrant is A

When we draw the given senerio we realise for A to be equidistant to both o and B it needs to be on the 2nd or 4th quadrant. only option in the second quadrant is A

hence A

There is an option which shows a point in 4th Quadrant (which one should notice even if one goes by quick observation like you have shown) however that is much closer to O than point B so that can't be the correct option.

Your approach is really good. exactly like one needed in any aptitude test. Overlooking option would be too bad for anyone taking GMAT
_________________

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

Re: In the rectangular coordinate system Point O has coordinates [#permalink]

Show Tags

26 Nov 2017, 03:09

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________