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In the rectangular coordinate system, are the points (r,s) [#permalink]
17 Apr 2010, 06:57
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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
Last edited by Bunuel on 16 Feb 2012, 21:22, edited 1 time in total.
Added the OA
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Re: GPrep - Coordinate [#permalink]
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=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]
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)^2So, 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: GPrep - Coordinate [#permalink]
21 Jan 2013, 05:09
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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=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: GPrep - Coordinate [#permalink]
19 Apr 2010, 02:48
Awesome. Thanks a bunch
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Re: GPrep - Coordinate [#permalink]
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)^2So, 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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Rectangular co-ordinate system [#permalink]
16 Feb 2012, 17:26
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 As the OA is not provided I think the answer should be C. I got this by substituting the values of r and s from statement 1 into statement 2. Any thoughts guys please?
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Re: Rectangular co-ordinate system [#permalink]
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: Rectangular co-ordinate system [#permalink]
16 Feb 2012, 21:24
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Re: In the rectangular coordinate system, are the points (r,s) [#permalink]
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]
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=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]
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]
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=ulikewise, from (1) s=1-r and from (2) s=1-v. so, r=vans: C
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Re: GPrep - Coordinate [#permalink]
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
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21 Jan 2013, 05:03
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