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Manager  Joined: 07 Sep 2010
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In the rectangular coordinate system Point O has coordinates  [#permalink]

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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.

A. (-2,6)
B. (0,4)
C. (2,-6)
D. (2,6)
E. (4,0)
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In the rectangular coordinate system Point O has coordinates  [#permalink]

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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)

Look at the diagram below: 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).

Hope it's clear.

Attachment: Area.png [ 11.71 KiB | Viewed 5931 times ]

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

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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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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.

Hope that helps a little.
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Re: In the rectangular coordinate system Point O has coordinates  [#permalink]

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There are three equidistant point:

(0, 4); (4,0); (-2;6)

Only the last one fulfills the area requirement!
Manager  Joined: 13 Aug 2012
Posts: 90
Re: In the rectangular coordinate system Point O has coordinates  [#permalink]

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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.

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

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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).

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

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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

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

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AmritaSarkar89 wrote:
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

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

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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)

Line joining OB is nothing but Y=x
A is equidistant from both the points means A lies on the perpendicular bisector of line OB.
Mid point of OB-(2,2), and A passes through this point.
So equation line perpendicular to line OB is
Y=-X+C(slpoe is -1 as Y=X slope is 1)
This line passes through (2,2)
2=-2+C
C=4
So line is
Y=-X+4
Options A,B and E satisfy this equation but For options B and E, Area OAB is 8
A:) Re: In the rectangular coordinate system Point O has coordinates   [#permalink] 11 Oct 2019, 09:51
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