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In the rectangular coordinate system, are the points (r,s)

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In the rectangular coordinate system, are the points (r,s) [#permalink]

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In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

(1) r + s = 1
(2) u = 1 - r and v = 1 - s
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Feb 2012, 21:22, edited 1 time in total.
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Re: GPrep - Coordinate [#permalink]

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In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

(1) r + s = 1

(2) u = 1 - r and v = 1 - s

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

Basically the question asks is \(\sqrt{r^2+s^2}=\sqrt{u^2+v^2}\) OR is \(r^2+s^2=u^2+v^2\)?

(1) \(r+s=1\), no info about \(u\) and \(v\);

(2) \(u=1-r\) and \(v=1-s\) --> substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1-r)^2+(1-s)^2=2-2(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=2-2(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=2-2(r+s)+ r^2+s^2\)? --> is \(r+s=1\)? We don't know that, so this statement is not sufficient.

(1)+(2) From (2) question became: is \(r+s=1\)? And (1) says that this is true. Thus taken together statements are sufficient to answer the question.

Answer: C.

Hope it helps.
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Re: GPrep - Coordinate [#permalink]

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New post 19 Apr 2010, 02:48
Awesome. Thanks a bunch :)
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Re: GPrep - Coordinate [#permalink]

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New post 01 Sep 2010, 19:20
Bunuel, I have a question:
How did you know that you had to express the equation in that way?
For example, I expressed (based on clue # 2) in this way:
\(r^2 + s^2 = (1-r)^2 + (1-s)^2\)
So, I obtain:
r + s = 1
The same as clue # 1. :(
How did you know that you had to do in the other way?

Thanks!
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Re: GPrep - Coordinate [#permalink]

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New post 01 Sep 2010, 19:48
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metallicafan wrote:
Bunuel, I have a question:
How did you know that you had to express the equation in that way?
For example, I expressed (based on clue # 2) in this way:
\(r^2 + s^2 = (1-r)^2 + (1-s)^2\)
So, I obtain:
r + s = 1
The same as clue # 1. :(
How did you know that you had to do in the other way?

Thanks!


Not sure I understand your question. But here is how I solved it:

The question asks: is \(r^2+s^2=u^2+v^2\)?

Then (2) says: \(u=1-r\) and \(v=1-s\). So now we can substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1-r)^2+(1-s)^2=2-2(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=2-2(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=2-2(r+s)+ r^2+s^2\)? --> is \(r+s=1\)? We don't know that, so this statement is not sufficient.

When combining: from (2) question became: is \(r+s=1\)? And (1) says that this is true. Thus taken together statements are sufficient to answer the question.

Hope it's clear.
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Re: Rectangular co-ordinate system [#permalink]

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New post 16 Feb 2012, 21:22
(r,s) and (u,v) will be equidistant from the origin when
r^2 + s^2 = u^2 + v^2

Using statement (1), r+s=1 gives us no information about u and v and so is insufficient.
Using statement (2), u = 1-r and v=1-s
=> r^2 + s^2 = (1-r)^2 + (1-s)^2
=> 2r + 2s - 2 = 0
or r + s = 1, which may or may not be true. Insufficient.

Combining (1) and (2) is clearly sufficient.

(C) is the answer.
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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]

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New post 02 Mar 2012, 00:52
I think it is a simple way to pick up values to solve this question because it is clear that each statement is not sufficient. For example;

for r=2, s=-1 we have u=-1, v=2 or for r=1, s=0 we have u=0, v=1 and so on. Therefore only if we know both statements, we can talk about the distance. So, the answer is C.
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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]

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New post 02 Mar 2012, 03:29
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ustureci wrote:
I think it is a simple way to pick up values to solve this question because it is clear that each statement is not sufficient. For example;

for r=2, s=-1 we have u=-1, v=2 or for r=1, s=0 we have u=0, v=1 and so on. Therefore only if we know both statements, we can talk about the distance. So, the answer is C.


This is not a good question for number picking. Notice that variables are not restricted to integers only, so r+s=1, u=1-r and v=1-s have infinitely many solutions for r, s, u and v.
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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]

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New post 31 Mar 2012, 10:21
best approch is imagine point to be on circumference of same circle.

Now radius of circle = use distance formula

so use equations in this logic. and get C 8-)
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rectangular coordinate system [#permalink]

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New post 01 Apr 2012, 09:26
question: \(r^2+s^2=u^2+v^2?\)

(a) insufficient because there's no info about u,v.
(b) insufficient. plug in numbers to see if it holds: find a 'yes' and then find a 'no'.

\((r,s)=(u,v)=(\frac{1}{2},\frac{1}{2})\) -------> 'yes' points are equidistant
\((r,s)=(0,0\), then \((u,v)=(1,1)\) -------> 'no' points are not equidistant

(c) together we can even prove it algebraically.
from (1) \(s=1-r\) and from (2) \(u=1-r\). so, \(s=u\)
likewise, from (1) \(s=1-r\) and from (2) \(s=1-v\). so, \(r=v\)

ans: C
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Re: GPrep - Coordinate [#permalink]

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New post 21 Jan 2013, 05:03
Bunuel wrote:
In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

(1) r + s = 1

(2) u = 1 - r and v = 1 - s

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

Basically the question asks is \(\sqrt{r^2+s^2}=\sqrt{u^2+v^2}\) OR is \(r^2+s^2=u^2+v^2\)?

(1) \(r+s=1\), no info about \(u\) and \(v\);

(2) \(u=1-r\) and \(v=1-s\) --> substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1-r)^2+(1-s)^2=2-2(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=2-2(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=2-2(r+s)+ r^2+s^2\)? --> is \(r+s=1\)? We don't know that, so this statement is not sufficient.

(1)+(2) From (2) question became: is \(r+s=1\)? And (1) says that this is true. Thus taken together statements are sufficient to answer the question.

Answer: C.

Hope it helps.


So the formula used here is different from the distance formula of square root of (x2-x1)^2 + (y2-y1)^2
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Re: GPrep - Coordinate [#permalink]

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New post 21 Jan 2013, 05:09
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Bunuel wrote:
In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

(1) r + s = 1

(2) u = 1 - r and v = 1 - s

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

Basically the question asks is \(\sqrt{r^2+s^2}=\sqrt{u^2+v^2}\) OR is \(r^2+s^2=u^2+v^2\)?

(1) \(r+s=1\), no info about \(u\) and \(v\);

(2) \(u=1-r\) and \(v=1-s\) --> substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1-r)^2+(1-s)^2=2-2(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=2-2(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=2-2(r+s)+ r^2+s^2\)? --> is \(r+s=1\)? We don't know that, so this statement is not sufficient.

(1)+(2) From (2) question became: is \(r+s=1\)? And (1) says that this is true. Thus taken together statements are sufficient to answer the question.

Answer: C.

Hope it helps.


So the formula used here is different from the distance formula of square root of (x2-x1)^2 + (y2-y1)^2


No it's not. The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\). Now, if one point is origin, coordinates (0, 0), then the formula can be simplified to: \(D=\sqrt{x^2+y^2}\).

Hope it's clear.
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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]

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New post 17 Sep 2014, 15:23
1) Not Suff as no info about u & v.
2) Not suff as 4 variables and 2 equations.

(1) and (2) combined:
From Statement (1), r =(1-s) = v by definition given in statement (2); and similarly s=(1-r)=u by definition given in statement (2).
Therefore s=u and r=v. Hence (r,s) and (u,v) represent same point and so have the same distance from origin. SUFF. Correct answer = C.
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In the rectangular coordinate system [#permalink]

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New post 23 Oct 2014, 06:44
In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

(1) r + s = 1

(2) u = 1-r and v = 1-s

Bunnel, can you please help me to get the quick solution to this problem ?
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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]

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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]

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Re: In the rectangular coordinate system, are the points (r,s)   [#permalink] 08 Dec 2015, 08:46
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