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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.

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! _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

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.

Re: Rectangular co-ordinate system [#permalink]
16 Feb 2012, 20:22

(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.

Re: In the rectangular coordinate system, are the points (r,s) [#permalink]
01 Mar 2012, 23:52

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.

Re: In the rectangular coordinate system, are the points (r,s) [#permalink]
02 Mar 2012, 02:29

Expert's post

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. _________________

rectangular coordinate system [#permalink]
01 Apr 2012, 08:26

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, \(r=v\)

Re: GPrep - Coordinate [#permalink]
21 Jan 2013, 04:03

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 _________________

Re: GPrep - Coordinate [#permalink]
21 Jan 2013, 04:09

1

This post received KUDOS

Expert's post

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: \(D=\sqrt{x^2+y^2}\).

Re: In the rectangular coordinate system, are the points (r,s) [#permalink]
17 Sep 2014, 14:23

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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