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

[Guest]

Please refer to the attachment.

The answer to the question is C.

Can someone explain why the answer is not B, as well as the underlying geometry principles this question is testing.

Can someone please highlight an actual numerical example when (r, s) and (1-r, 1-s) are both equidistant from the origin?

Thank you!

[anadi]

Distance from origin is x^2+y^2

r,s distance is r^2+s^2

(1-r), (1-s) distance is (1-r)^2+(1-s)^2 = 1+r^2-2r+1+s^2-2s = r^2+s^2 + 2(1-r-s) which is same as r,s distance if r+s = 1, since in that case, 1-r-s = 0.

Numerical example,

2,2 is not at the same ditance from origin as 1,1 (which is 2-1, 2-1).
But if we take a example where r+s = 1

say 1/2, 3/2 then 1-1/2, 1-3/2 (effectively 3/2, 1/2) is same distance as 1/2,3/2.

[Guest]

Your post was helpful.

thank you.

[rumni]

Can someone from Manhattan gmat explain this one?

[RonPurewal]

Can someone from Manhattan gmat explain this one?

--

statement 1:
obviously insufficient, as no information at all is provided about u or v.

--

statement 2:
if you like this sort of algebra enough to perform it in under two minutes, then you can probably work it out; post back if you want us to show all the details.
if you don't like doing this sort of algebra, you can pick numbers with the purpose of finding a "yes" and a "no" to the prompt question.

the easiest way to get a "yes" is to select numbers so that the coordinates are all the same.
since u = 1 - r, you can make u and r equal by setting both of them equal to 1/2.
same goes for v and s.
therefore, (1/2, 1/2) and (1/2, 1/2) give a YES to the prompt question.

to get a "no", put one point AT the origin, so that the distance in question, for that point, becomes zero. that way, as long as the other point turns out to be anywhere other than at the origin, you're good.
if you set (r, s) to be (0, 0), then (u, v) winds up at (1, 1). these points are not equidistant from the origin.

--

together
first, a preliminary observation:
if the 2 coordinates of one point have the same magnitude (absolute value), IN SOME ORDER, as the 2 coordinates of another point, then those 2 points are equidistant from the origin.**
there's no need to calculate the actual distances in this case.

since this is ridiculously awkward to state in words, here's an example:
consider the point (4, 2). the statement above means that any other point with a "4" and a "2" as coordinates, no matter in what order or whether positive/negative, will be the same distance from the origin as is (4, 2) itself.
in other words, (4, -2), (-4, 2), (-4, -2), (2, 4), (-2, 4), (2, -4), and (-2, -4) are all the same distance from the origin as is (4, 2).
you don't have to use the distance formula to figure this out: just realize that each of them is 2 units either vertically or horizontally, and then 4 units in the orthogonal direction, away from the origin.

two ways to approach this:

(1) algebra / substitution:
statement 1 gives s = 1 - r. since u is also equal to 1 - r, we have s = u.
also, v = 1 - s, so, substituting the above, v = 1 - (1 - r) --> v = r.
since s = u and v = r, the above preliminary observation guarantees that (r, s) and (u, v) will be equidistant from the origin.

(2) plug in numbers:
if you plug in random r's, the pattern will become ridiculously obvious very quickly. again, as i say in almost every strategy post, you should not hesitate to plug in numbers; if the algebra is not yielding to your smooth talk right away, go for the number plugging instead.
try r = 0 --> this gives (r, s) = (0, 1) and (u, v) = (1, 0). equidistant.
try r = 0.5 --> this gives (0.5, 0.5) and (0.5, 0.5). equidistant.
try r = 10 --> this gives (10, -9) and (-9, 10). equidistant.
try r = -3.4 --> this gives (-3.4, 4.4) and (4.4, -3.4). equidistant.
you can see what's happening. (you should be convinced by the time you've done the first three, but, if you're not sure whether decimals would spoil the action, go ahead and plug in something like the fourth.)
sufficient.

answer = c

[nlaszchuk]

This question is for RPurewall regarding your below post.
I can understand substituting 1/2 for r, s as 1/2 + 1/2 = 1 which satisfies option 1).
However, I'm not sure how you justify setting r and s equal to 0 when 0 + 0 cannot equal 1... Am I losing it here??

I am also rather confused about the explanations the posts above that use squares to solve the equation.

... 'Elp!

Can someone from Manhattan gmat explain this one?

--

statement 1:
obviously insufficient, as no information at all is provided about u or v.

--

statement 2:
if you like this sort of algebra enough to perform it in under two minutes, then you can probably work it out; post back if you want us to show all the details.
if you don't like doing this sort of algebra, you can pick numbers with the purpose of finding a "yes" and a "no" to the prompt question.

the easiest way to get a "yes" is to select numbers so that the coordinates are all the same.
since u = 1 - r, you can make u and r equal by setting both of them equal to 1/2.
same goes for v and s.
therefore, (1/2, 1/2) and (1/2, 1/2) give a YES to the prompt question.

to get a "no", put one point AT the origin, so that the distance in question, for that point, becomes zero. that way, as long as the other point turns out to be anywhere other than at the origin, you're good.
if you set (r, s) to be (0, 0), then (u, v) winds up at (1, 1). these points are not equidistant from the origin.

--

together
first, a preliminary observation:
if the 2 coordinates of one point have the same magnitude (absolute value), IN SOME ORDER, as the 2 coordinates of another point, then those 2 points are equidistant from the origin.**
there's no need to calculate the actual distances in this case.

since this is ridiculously awkward to state in words, here's an example:
consider the point (4, 2). the statement above means that any other point with a "4" and a "2" as coordinates, no matter in what order or whether positive/negative, will be the same distance from the origin as is (4, 2) itself.
in other words, (4, -2), (-4, 2), (-4, -2), (2, 4), (-2, 4), (2, -4), and (-2, -4) are all the same distance from the origin as is (4, 2).
you don't have to use the distance formula to figure this out: just realize that each of them is 2 units either vertically or horizontally, and then 4 units in the orthogonal direction, away from the origin.

two ways to approach this:

(1) algebra / substitution:
statement 1 gives s = 1 - r. since u is also equal to 1 - r, we have s = u.
also, v = 1 - s, so, substituting the above, v = 1 - (1 - r) --> v = r.
since s = u and v = r, the above preliminary observation guarantees that (r, s) and (u, v) will be equidistant from the origin.

(2) plug in numbers:
if you plug in random r's, the pattern will become ridiculously obvious very quickly. again, as i say in almost every strategy post, you should not hesitate to plug in numbers; if the algebra is not yielding to your smooth talk right away, go for the number plugging instead.
try r = 0 --> this gives (r, s) = (0, 1) and (u, v) = (1, 0). equidistant.
try r = 0.5 --> this gives (0.5, 0.5) and (0.5, 0.5). equidistant.
try r = 10 --> this gives (10, -9) and (-9, 10). equidistant.
try r = -3.4 --> this gives (-3.4, 4.4) and (4.4, -3.4). equidistant.
you can see what's happening. (you should be convinced by the time you've done the first three, but, if you're not sure whether decimals would spoil the action, go ahead and plug in something like the fourth.)
sufficient.

answer = c

[RonPurewal]

I'm not sure how you justify setting r and s equal to 0 when 0 + 0 cannot equal 1... Am I losing it here??

that was done in the treatment of statement (2) alone.
the requirement that r + s = 1 is only a part of statement (1), and so is irrelevant to the consideration of statement (2).

I am also rather confused about the explanations the posts above that use squares to solve the equation.

the squares are an essential part of the way that distance is defined in an x-y coordinate system.

given two points in an x-y coordinate system, the distance between those points is defined as
square root of ( (distance between x's)^2 + (distance between y's)^2 ).

are you aware of this fact?