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

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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chetan2u Bunuel VeritasKarishma

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.

Awesome. Thanks a bunch

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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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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(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.

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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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

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

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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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.

Yes, you are correct in the present state.
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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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Hi GyanOne : you have mentioned that after solving statement 2 we get 2r+2s = 1 , which is equivalent to r+s =1. Statement 2 gives us the same thing as statement 1.
Combining the two statement how do we get sufficiency? I see both the statements are resulting equivalent fact. Can you please help my understanding ?

Thanks

Why can't we use the mid-point formula in this question i.e. ((r+u)/2) = 0 and ((s+v)/2) = 0 (if the points are equidistant, this should be true) and from B, it isn't the case i.e. r+u=1, thus, clear NO and B is the answer.

KarishmaB

Mid point formula gives us the mid point of a line segment.

So if two end points of a line segment are the points (r,s) and (u,v), the mid point formula will give the point lying at the middle of the line segment joining these points. But the origin may not be this point. We are not given that the three points are collinear.
Two points can be equidistant from a third point even if the third point does not lie at the centre of the line joining them.
Every point on the perpendicular bisector of that line segment will be equidistant from the two end points.
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Got it, thank you KarishmaB