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In the rectangular coordinate system, are the points (r,s) [#permalink]
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Updated on: 16 Feb 2012, 21:22
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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
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Originally posted by calvinhobbes on 17 Apr 2010, 06:57.
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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17 Apr 2010, 07:31
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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) \(u=1r\) and \(v=1s\) > substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1r)^2+(1s)^2=22(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=22(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=22(r+s)+ r^2+s^2\)? > 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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19 Apr 2010, 02:48
Awesome. Thanks a bunch



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Re: GPrep  Coordinate [#permalink]
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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: \(r^2 + s^2 = (1r)^2 + (1s)^2\) 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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01 Sep 2010, 19:48
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: \(r^2 + s^2 = (1r)^2 + (1s)^2\) 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: \(u=1r\) and \(v=1s\). So now we can substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1r)^2+(1s)^2=22(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=22(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=22(r+s)+ r^2+s^2\)? > 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 coordinate system [#permalink]
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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 = 1r and v=1s => r^2 + s^2 = (1r)^2 + (1s)^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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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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02 Mar 2012, 03:29
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=1r and v=1s 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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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



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rectangular coordinate system [#permalink]
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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) \(s=1r\) and from (2) \(u=1r\). so, \(s=u\) likewise, from (1) \(s=1r\) and from (2) \(s=1v\). so, \(r=v\) ans: C
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Re: GPrep  Coordinate [#permalink]
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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) \(u=1r\) and \(v=1s\) > substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1r)^2+(1s)^2=22(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=22(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=22(r+s)+ r^2+s^2\)? > 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 (x2x1)^2 + (y2y1)^2
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Re: GPrep  Coordinate [#permalink]
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21 Jan 2013, 05:09
fozzzy wrote: 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) \(u=1r\) and \(v=1s\) > substitute \(u\) and \(v\) and express RHS using \(r\) and \(s\) to see what we get: \(RHS=u^2+v^2=(1r)^2+(1s)^2=22(r+s)+ r^2+s^2\). So we have that \(RHS=u^2+v^2=22(r+s)+ r^2+s^2\) and thus the question becomes: is \(r^2+s^2=22(r+s)+ r^2+s^2\)? > 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 (x2x1)^2 + (y2y1)^2 No it's not. The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). 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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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 =(1s) = v by definition given in statement (2); and similarly s=(1r)=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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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]
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