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# O is the origin of the coordinate system above, and A lies in the 2nd

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O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 02:20
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O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 05:22
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

Let the coordinates of A = A(x,y)

OAB is a right triangle with right angle at A

Per statement 1, OA=5 ---> A can be either -3,4 or -4,3. Thus this statement is not sufficient.

Per statement 2, AB=5. Clearly not sufficient.

Combining, you get OA=AB=5 giving you a definite coordinate of A by virtue of OAB becoming an isosceles right triangle.

Manager
Joined: 17 Oct 2013
Posts: 56
Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 06:01
Engr2012 wrote:
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

Let the coordinates of A = A(x,y)

OAB is a right triangle with right angle at A

Per statement 1, OA=5 ---> A can be either -3,4 or -4,3. Thus this statement is not sufficient.

Per statement 2, AB=5. Clearly not sufficient.

Combining, you get OA=AB=5 giving you a definite coordinate of A by virtue of OAB becoming an isosceles right triangle.

I tired it different way. Since A is 90, if I put a perpendicular from A to OB, lets call that point D. Now ODA will be 90 and since we are bisecting angle A, it will be 45. Now we have three angles.

D=90
A=45
O=45

using 45:45:90 we have 1:1:sqrt2, now we know OA is 5, we can easily check the length of OD and AD hence points of A.

What is wrong in this calculation.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 06:04
viksingh15 wrote:
Engr2012 wrote:
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

Let the coordinates of A = A(x,y)

OAB is a right triangle with right angle at A

Per statement 1, OA=5 ---> A can be either -3,4 or -4,3. Thus this statement is not sufficient.

Per statement 2, AB=5. Clearly not sufficient.

Combining, you get OA=AB=5 giving you a definite coordinate of A by virtue of OAB becoming an isosceles right triangle.

I tired it different way. Since A is 90, if I put a perpendicular from A to OB, lets call that point D. Now ODA will be 90 and since we are bisecting angle A, it will be 45. Now we have three angles.

D=90
A=45
O=45

using 45:45:90 we have 1:1:sqrt2, now we know OA is 5, we can easily check the length of OD and AD hence points of A.

What is wrong in this calculation.

I dont undertsand how you are assuming that the perpendicular drawn from A to OB will bisect angle A. Is it after a particular statement or after you cmobine both statements?

If you assume the above after combining the statements, then yes you are correct but if you are assuming the above after one of the statements individually then you are wrong.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 06:10
If I remember correctly, when you draw a perpendicular from 90 degree to hypotenuse, it bisects 90 degree to 45-45.

I may be wrong.
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O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 06:12
viksingh15 wrote:
If I remember correctly, when you draw a perpendicular from 90 degree to hypotenuse, it bisects 90 degree to 45-45.

I may be wrong.

But who said OAB is an isosceled right triangle ? You are only able to say that it is an isosceles right triangle after combining the 2 statements. This is what I have mentioned in my solution above.

What should the solution be according to you?
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 08:02
Engr2012 wrote:
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

Let the coordinates of A = A(x,y)

OAB is a right triangle with right angle at A

Per statement 1, OA=5 ---> A can be either -3,4 or -4,3. Thus this statement is not sufficient.

Per statement 2, AB=5. Clearly not sufficient.

Combining, you get OA=AB=5 giving you a definite coordinate of A by virtue of OAB becoming an isosceles right triangle.

Hi,

In statement one, my inference was different as I assumed that there might be values other than -3, 4 and -4, 3 that might satisfy the equation. The values could be in fractions right?
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 08:05
1
longfellow wrote:
Engr2012 wrote:
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

Let the coordinates of A = A(x,y)

OAB is a right triangle with right angle at A

Per statement 1, OA=5 ---> A can be either -3,4 or -4,3. Thus this statement is not sufficient.

Per statement 2, AB=5. Clearly not sufficient.

Combining, you get OA=AB=5 giving you a definite coordinate of A by virtue of OAB becoming an isosceles right triangle.

Hi,

In statement one, my inference was different as I assumed that there might be values other than -3, 4 and -4, 3 that might satisfy the equation. The values could be in fractions right?

Absolutely. The choices I mentioned led me to 2 different options quite easily and thus I was able to provde insufficiency of the statement. Your point could even have been $$(-2, \sqrt{21})$$.

Hope this helps.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 08:35
Engr2012 wrote:
longfellow wrote:
Engr2012 wrote:
Let the coordinates of A = A(x,y)

OAB is a right triangle with right angle at A

Per statement 1, OA=5 ---> A can be either -3,4 or -4,3. Thus this statement is not sufficient.

Per statement 2, AB=5. Clearly not sufficient.

Combining, you get OA=AB=5 giving you a definite coordinate of A by virtue of OAB becoming an isosceles right triangle.

Hi,

In statement one, my inference was different as I assumed that there might be values other than -3, 4 and -4, 3 that might satisfy the equation. The values could be in fractions right?

Absolutely. The choices I mentioned led me to 2 different options quite easily and thus I was able to provde insufficiency of the statement. Your point could even have been $$(-2, \sqrt{21})$$.

Hope this helps.

It seems I am missing something here. How is it an isosceles triangle?? Having a really bad day today.
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O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 08:37
longfellow wrote:

It seems I am missing something here. How is it an isosceles triangle?? Having a really bad day today.

No problem. Look below.

In triangle OAB, OA=AB=5 (after you combine both the statements). Thus making triangle OAB as isosceles right triangle with angles OBA and AOB = 45 degrees each.

Hope this helps.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 08:43
Engr2012 wrote:
longfellow wrote:

It seems I am missing something here. How is it an isosceles triangle?? Having a really bad day today.

No problem. Look below.

In triangle OAB, OA=AB=5 (after you combine both the statements). Thus making triangle OAB as isosceles right triangle with angles OBA and AOB = 45 degrees each.

Hope this helps.

Thanks Engr2012. I got that as soon as I posted the question.
Cant seem to get my brain to work today. Got my GMAT in 10 days and was feeling good and ready for it till now but today I seem to have gone blank.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Oct 2015, 09:25
1
longfellow wrote:
Engr2012 wrote:
longfellow wrote:

It seems I am missing something here. How is it an isosceles triangle?? Having a really bad day today.

No problem. Look below.

In triangle OAB, OA=AB=5 (after you combine both the statements). Thus making triangle OAB as isosceles right triangle with angles OBA and AOB = 45 degrees each.

Hope this helps.

Thanks Engr2012. I got that as soon as I posted the question.
Cant seem to get my brain to work today. Got my GMAT in 10 days and was feeling good and ready for it till now but today I seem to have gone blank.

This just shows that you are trying too hard and that you should give yourself a few hours off from GMAT related work. Come back after a break and you will feel refreshed and ready to go for the home stretch. All the best for your GMAT.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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10 Apr 2016, 19:36
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

each alone is not sufficient.
1+2
we have 45-45-90 triangle. BO is 5*sqrt(2).. y coordinate will thus be 2.5*sqrt(2).
now..we can draw a perpendicular (to y axis) bisector. we get 2 separate 45-45-90 small triangles.
we know the hypotenuse 5, we know the leg 2.5*sqrt(2). we can find the length of the second leg. so C should be sufficient.
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O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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28 Jun 2016, 10:44
Hi everyone

How its not D? We know that the length of AO and AB is the same. We know it from 1) OR 2) BECAUSE we know the 90degree angle. This guarantees that the two lines are mirror images of each other, if one is 5 the other must be 5 too, it can't be 6 or 9 or whatever!

So by calculating the area of the triangle ABO we get its high from y axis to A so we get the X coordinate. From there we use triangles again to find y coordinate. As long as we know the length of AO or AB we have enough info to answer! Im 100% sure lol (or 100% wrong)

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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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29 Jun 2016, 00:42
iliavko wrote:
Hi everyone

How its not D? We know that the length of AO and AB is the same. We know it from 1) OR 2) BECAUSE we know the 90degree angle. This guarantees that the two lines are mirror images of each other, if one is 5 the other must be 5 too, it can't be 6 or 9 or whatever!

So by calculating the area of the triangle ABO we get its high from y axis to A so we get the X coordinate. From there we use triangles again to find y coordinate. As long as we know the length of AO or AB we have enough info to answer! Im 100% sure lol (or 100% wrong)

No, that's not correct. From (1) as well as from (2) we have a right triangle with one leg equal to 5 units. We know nothing else. The other leg can be 1, 10, 1,000,000, basically any positive number.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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29 Jun 2016, 01:57
Hi, Bunuel,

Does it mean that the lines could be something like this?
Attachments

angle.jpg [ 7.84 KiB | Viewed 1338 times ]

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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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29 Jun 2016, 21:59
iliavko wrote:
Hi, Bunuel,

Does it mean that the lines could be something like this?

_______________________
Correct.
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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08 Aug 2017, 03:32
reto wrote:
Attachment:
T6133.png

O is the origin of the coordinate system above, and A lies in the 2nd quadrant. If angle OAB is a right angle, what are the coordinates of point A?

(1) The length of line segment OA is 5.
(2) The length of line segment AB is 5.

Given : O is the origin of coordinate system.
A lies in 2nd quadrant. /_OAB = 90 deg.

DS: Coordinates of A.

Statement 1: Length of OA = 5 and /_OAB = 90 deg. But OA can be at any angle with x -axis in the 2nd quadrant as length of AB is unknown
NOT SUFFICIENT

STatement 2 : Length AB = 5 and /_OAB = 90 deg. But AB can be at any angle with y -axis in the 2nd quadrant as length of OA is unknown

Combined : Length of AB = 5 . Now this fix the position of OA so that AB is fixed at 5 and the angle /OAB = 90 deg.
So OA = AB = 5 make OAB an isosceles triangle with /_AOB = /_ABO = 45 deg.
SO, /_ made by OA with x-axia is 45 deg.

Thus coordinate A is (-5 Cos 45deg, 5 sin 45deg) = (-5/\sqrt{2}, 5/\sqrt{2})

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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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12 Aug 2017, 12:28
1
My 2 cents:
Above explanation is sufficient - i thought this diagram will help make sense of it...
Attachment:

my 2 cents.png [ 19.46 KiB | Viewed 641 times ]
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Re: O is the origin of the coordinate system above, and A lies in the 2nd [#permalink]

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15 Aug 2017, 00:55

I am not sure how the answer is C. Should'nt the answer be E?

Using C. We know that the the triangle is an isosceles right triangle with lengths 5:5:5sqrt2 and point A is on the 2 quadrant.

How can we determine the points exactly of A & B ?

Re: O is the origin of the coordinate system above, and A lies in the 2nd   [#permalink] 15 Aug 2017, 00:55

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