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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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jbpayne wrote:
Thats what I thought, but the answer is B, 1. Anyone able to figure this one out?


There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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there is no need for much calculation,
simply switch their X and Y

P and Q are always perpendicular and same distance from O

OP is perpendicular to OQ means you rotate OQ 90 degrees, you will
get OP, in terms of X,Y value, that is just to switch them and flip the sign
to have (-Y,X)
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....

Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=-1
Since m1= -1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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The answer is B.

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle \(\triangle NOP\). We are given the co-ordinates of P as \((-\sqrt{3}, 1)\). i.e. \(ON=\sqrt{3}\) and \(PN=1\). Also we know angle \(PNO=90\textdegree\). If you notice this is a \(30\textdegree-60\textdegree-90\textdegree\)triangle with sides \(1-sqrt3-2\) and angle \(NOP=30\textdegree\).

Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis R. Now we have a right angle triangle \(\triangle QOR\). We know angle \(QRO=90\textdegree\).

Now
angle NOP + angle POQ (= 90 -- given) + angle QOR = 180

\(\Rightarrow 30\textdegree+90\textdegree+angleQOR=180\)

\(\Rightarrow angleQOR=60\)

If you notice \(\triangle QOR\) is also a \(30\textdegree-60\textdegree-90\textdegree\)triangle with sides \(1-sqrt3-2\) and angle \(OQR=30\textdegree\)

The side opposite to this angle is OR = 1.

Hence answer is B.
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The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way:

Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines?
Attachment:
Ques1.jpg
Ques1.jpg [ 7.87 KiB | Viewed 184771 times ]

Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line \(= \sqrt{3}\) cm and blue line = 1 cm

So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y co-ordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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bsd_lover wrote:
Could you please elaborate the bolded part ?

sondenso wrote:
There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1


I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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sondenso wrote:
bsd_lover wrote:
Could you please elaborate the bolded part ?

sondenso wrote:
There are more than one way to solve this. I like this way

The line PO has slope =-1/√3
The line QO has slope = t/s
PO and QO is perpendicular so [-1/√3]*t/s = -1 ==> t/s =√3 or √3/1 so s = 1


I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1




From the picture t^2 +s^2 = 4(radius = 2)
By substituting t =√3s we get s=1
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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Check another approach for this problem.
Attachments

ps_ans1.JPG
ps_ans1.JPG [ 25.21 KiB | Viewed 319190 times ]

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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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Since OP and OQ are the radii of the circle, and we know P has co-ordinates (-\sqrt{3}, 1), from the distance formula we have OP = OQ = 2. Therefore, s^2+t^2=4. -> (1)
Also from the right triangle, we know the length of PQ=2\sqrt{2}.
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

From (1) and (2) you can solve for s.
Hope this helps.
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I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.
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Attachment:
a.jpg
a.jpg [ 12.91 KiB | Viewed 316150 times ]


Line OP => y=mx => 1 = \(-sqrt{3} * m\)

=> slope of OP = \(\frac{-1}{sqrt{3}}\)

Since OP is perpendicular to OQ => m1*m2 = -1
=> slope of OQ = \(sqrt{3}\)

equation of line OQ => y = \(sqrt{3}x\)

since s and t lie on this line t= \(sqrt{3}s\)

also radius of circle = \(sqrt{ 1^2+ 3}\) = 2 = \(sqrt{ s^2+ t^2}\)

=> \(4 = s^2+ t^2 = s^2 + 3s^2 = = 4s^2\)
=> s=1
----------------------------------------------------------------------

Attachment:
b.jpg
b.jpg [ 13.36 KiB | Viewed 315555 times ]


Using Similar triangles.
In triangle APO , angle APO + AOP = 90 ---------------------1
In triangle BOQ , angle BOQ + BQO= 90 ---------------------2

Also SINCE angle POQ = 90 as given.

angle AOP + angle BOQ = 90 ------------------3

Using equations 1 and 3 we get Angle APO = Angle BQO -----------4

using equation 2 ,3, 4
we get Angle APO = Angle BQO and AOP = OQB

Since hypotenuse is common => both the triangle are congruent

=> base of triangle APO = height of triangle AQO and vise versa.

Thus s = y co-ordinate of P = 1

Originally posted by gurpreetsingh on 09 Sep 2010, 16:46.
Last edited by gurpreetsingh on 09 Sep 2010, 17:00, edited 1 time in total.
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tonebreeze: As requested, I am giving you my take on this problem (though I guess you have already got some alternative solutions that worked for you)

I would suggest you first go through the explanation at the link given below for another question. If you understand that, solving this question is just a matter of seconds by looking at the diagram without using any formulas.
https://gmatclub.com/forum/coordinate-plane-90772.html#p807400

P is (\(-\sqrt{3}\), 1). O is the center of the circle at (0,0). Also OP = OQ. When OP is turned 90 degrees to give OQ, the length of the x and y co-ordinates will get interchanged (both x and y co-ordinates will be positive in the first quadrant). Hence s, the x co-ordinate of Q will be 1 and y co-ordinate of Q will be \(\sqrt{3}\).
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Note that this is not a reflection problem. The angle POQ is given as = 90, but angle POY is not the same as angle QOY.

To solve, set the product of the slopes of OP and OQ as -1
=> -1/sqrt(3) * t/s =-1
=> t = s*sqrt(3)

Also OQ=OP
=> s^2 + t^2 = 4
=> s^2 + 3*s^2 = 4
=> s^2 = 1
=> s = 1

Option (B)
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This is not a question on reflection. This is a question on finding the coordinates of a given point within some stated conditions. It would have been a reflection question if angle POY was the same as angle QOY. It is not, so this is not a reflection question. Solve it according to the rules explained in my last post.

I took the product of the slopes as -1 because if two lines are perpendicular, the product of their slopes must be -1.
OP and OQ are clearly perpendicular as given in the figure, so their slopes must have a product of -1.
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.


In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\);
Point P in II quadrant \((-\sqrt{3}, 1)\);
Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP);
Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).


Is it the case of reflection in Y axis ?
I doubt that but still want to clarify..
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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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ygdrasil24 wrote:
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.


In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant \((1, \sqrt{3})\);
Point P in II quadrant \((-\sqrt{3}, 1)\);
Point T in III quadrant \((-1, -\sqrt{3})\) (OT perpendicular to OP);
Point R in IV quadrant \((\sqrt{3},-1)\) (OR perpendicular to OQ).


Is it the case of reflection in Y axis ?
I doubt that but still want to clarify..


No. \((-\sqrt{3}, 1)\) is not a reflection of \((1, \sqrt{3})\) around the Y-axis. The reflection of \((1, \sqrt{3})\) around the Y-axis is \((-1, \sqrt{3})\).
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Hello all, I just got this question in my GMATprep practice test, and came looking here for an explanation afterwards.

I understand the math in the comments above, and why the answer is what it is. What concerns me, is that the image is not drawn to scale.
It did not say the image was not drawn to scale, and this is an official question, so why could I not assume the image was drawn to scale?

(When one has time to look at the question properly it is obvious that it is not drawn to scale - but I have up to now just been assuming that geometry questions are drawn to scale unless clearly indicated that they are not.)
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