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# In the rectangular coordinate system, are the points (r,s) and (u,v)

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

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Updated on: 06 Feb 2019, 04:09
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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

Originally posted by calvinhobbes on 17 Apr 2010, 06:57.
Last edited by Bunuel on 06 Feb 2019, 04:09, edited 2 times in total.
Renamed the topic.
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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#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) $$u=1-r$$ 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.

Hope it helps.
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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#permalink]

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01 Sep 2010, 19:48
2
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: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#permalink]

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21 Jan 2013, 05:09
1
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.

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) and (u,v)  [#permalink]

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07 Jun 2019, 00:04
1
sakshamchhabra wrote:

Greetings Experts,

while trying to figure out a shorter approach for this question, I noticed -

St 1. r + s = 1 -----> r = 1 - s OR s = 1 - r

St 2. u = 1 - r and v = 1 - s

from the above statements, we can deduce ---> u = s and v = r

Hence, the points will definitely be equidistant.

Please correct me if I am wrong.

Yes, you are correct in the present state.
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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#permalink]

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07 Jun 2019, 02:01
1
sakshamchhabra wrote:

Greetings Experts,

while trying to figure out a shorter approach for this question, I noticed -

St 1. r + s = 1 -----> r = 1 - s OR s = 1 - r

St 2. u = 1 - r and v = 1 - s

from the above statements, we can deduce ---> u = s and v = r

Hence, the points will definitely be equidistant.

Please correct me if I am wrong.

Yes, your logic works and it's great!
Note that r = 1 - s AND s = 1 - r
Since r and s add up to 1, whatever the value of r, value of s will be complementary to that. So r will be 1 - s and s will be 1 - r.
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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#permalink]

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19 Apr 2010, 02:48
Awesome. Thanks a bunch
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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#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:
$$r^2 + s^2 = (1-r)^2 + (1-s)^2$$
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: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#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 = 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.

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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#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) and (u,v)  [#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=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) and (u,v)  [#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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Re: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#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) $$s=1-r$$ 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: In the rectangular coordinate system, are the points (r,s) and (u,v)  [#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) $$u=1-r$$ 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.

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

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06 Jun 2019, 11:08

Greetings Experts,

while trying to figure out a shorter approach for this question, I noticed -

St 1. r + s = 1 -----> r = 1 - s OR s = 1 - r

St 2. u = 1 - r and v = 1 - s

from the above statements, we can deduce ---> u = s and v = r

Hence, the points will definitely be equidistant.

Please correct me if I am wrong.
Re: In the rectangular coordinate system, are the points (r,s) and (u,v)   [#permalink] 06 Jun 2019, 11:08
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