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

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11 Jun 2013, 06:56

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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 3363 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]

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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]

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11 Jun 2013, 08:52

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

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11 Jun 2013, 15:20

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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]

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02 Aug 2013, 09:18

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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]

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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]

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20 Sep 2014, 03:58

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

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01 Aug 2016, 03:26

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).

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

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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
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