In a rectangular coordinate system, are the points (r,s) & (u,v) equidistant for the origin
I r+s=1
II u=1-r & v=1-s
Answer is C
Can someone please explain?
Thanks.
Coordinate System
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- codesnooker
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It's a bit tricky question and based on properties of triangle.
Concept: The distance between two point will we same on interchanging the values of x-coordinate and y-coordinate of a point and keeping fixed the other point.
That is, suppose we have three points A(0, 0), B(x, y) and C(y, x).
So according the concept stated above, the distance between AB and AC will be same. (It can be prove through Pythagoras theorem).
So in the current question, if we can prove that r = v and s = u then we can say that (r, s) and (u, v) are equidistant from the origin.
According to statement (1), we can't get any information regarding the point (u, v). Hence ALONE INSUFFICIENT. Therefore, (A) and (D) are out.
According to statement (2), we can know the relationship between u and r, s and v, but still we can't prove that r = v and s = u. Hence ALONE INSUFFICIENT. Therefore, (B) is also out.
Now lets take both the statements together.
Statement (1) states that, r + s = 1.
that is, r = 1 - s
Now according to statement (2),
v = 1 - s. Therefore, r = v.
Similarly, s = u.
So, we have proved that s = u and r = v, hence (u, v) and (r, s) are equidistant from the origin.
Hope this helps...
Concept: The distance between two point will we same on interchanging the values of x-coordinate and y-coordinate of a point and keeping fixed the other point.
That is, suppose we have three points A(0, 0), B(x, y) and C(y, x).
So according the concept stated above, the distance between AB and AC will be same. (It can be prove through Pythagoras theorem).
So in the current question, if we can prove that r = v and s = u then we can say that (r, s) and (u, v) are equidistant from the origin.
According to statement (1), we can't get any information regarding the point (u, v). Hence ALONE INSUFFICIENT. Therefore, (A) and (D) are out.
According to statement (2), we can know the relationship between u and r, s and v, but still we can't prove that r = v and s = u. Hence ALONE INSUFFICIENT. Therefore, (B) is also out.
Now lets take both the statements together.
Statement (1) states that, r + s = 1.
that is, r = 1 - s
Now according to statement (2),
v = 1 - s. Therefore, r = v.
Similarly, s = u.
So, we have proved that s = u and r = v, hence (u, v) and (r, s) are equidistant from the origin.
Hope this helps...
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I guess we can use the distance formula : For two points r,s and u,v to be equidistant from the origin
r^2+s^2 = u^2+v^2
Stmnt 1 - clearly not sufficient
Stmnt 2 - not sufficient either
Combining 1 and 2 we see that s^2-2s+1 = s^2 -2s+1 and hence sufficient
r^2+s^2 = u^2+v^2
Stmnt 1 - clearly not sufficient
Stmnt 2 - not sufficient either
Combining 1 and 2 we see that s^2-2s+1 = s^2 -2s+1 and hence sufficient
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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
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
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rosh26 wrote:In a rectangular coordinate system, are the points (r,s) & (u,v) equidistant for the origin
I r+s=1
II u=1-r & v=1-s
Answer is C
Can someone please explain?
Thanks.
Is r^2 + s^2 = u^2 + v^2?
[1] r + s = 1, this is insufficient.
[2] u = 1 - r and v = 1 - s, this makes r^2 + s^2 = 1 - 2 u + u^2 + 1 - 2 v + v^2 = u^2 + v^2, if and only if u + v = 1; who knows? Insufficient
Taken together
If r + s = 1, then u + v = 1 - r + 1 - s = 2 - (r + s) = 2 - 1 = 1, and then, [spoiler]r^2 + s^2 = u^2 + v^2. Sufficient
C[/spoiler]
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what "distance" formula is it thats' being applied here? by squaring the "r" coordinate - how are you finding distance between r and s?pbanavara wrote:I guess we can use the distance formula : For two points r,s and u,v to be equidistant from the origin
r^2+s^2 = u^2+v^2
Stmnt 1 - clearly not sufficient
Stmnt 2 - not sufficient either
Combining 1 and 2 we see that s^2-2s+1 = s^2 -2s+1 and hence sufficient
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see here for information:missrochelle wrote:
what "distance" formula is it thats' being applied here? by squaring the "r" coordinate - how are you finding distance between r and s?
https://www.purplemath.com/modules/distform.htm
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Ron,
Here is my basic question :
Option II u=1-r & v=1-s --? Eq 1
Question is : are the points (r,s) & (u,v) equidistant for the origin . Does that mean the midpoint for both of these need to be origin ??
If so { ( r+u) /2 , (s+v)/2 } --> Eq 2
By subtituting eq1 in eq2
{ (r+1-r)/2 , ( 1-s+s)/2 ) } ==> {1/2,1/2}
Since its not equal to origin cant i conclude that this answer the Question as " NO" , so i choose B .
What is that i missed here in this process. Appreciate your assistance .
Here is my basic question :
Option II u=1-r & v=1-s --? Eq 1
Question is : are the points (r,s) & (u,v) equidistant for the origin . Does that mean the midpoint for both of these need to be origin ??
If so { ( r+u) /2 , (s+v)/2 } --> Eq 2
By subtituting eq1 in eq2
{ (r+1-r)/2 , ( 1-s+s)/2 ) } ==> {1/2,1/2}
Since its not equal to origin cant i conclude that this answer the Question as " NO" , so i choose B .
What is that i missed here in this process. Appreciate your assistance .
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no. equidistant just means "same distance away from". they don't have to be in opposite directions.sandysai wrote: Question is : are the points (r,s) & (u,v) equidistant for the origin . Does that mean the midpoint for both of these need to be origin ??
for instance, (1, 0) and (0, 1) are equidistant from the origin, since each of them is 1 unit away from the origin.
--
did you check out this post earlier on this thread?
https://www.beatthegmat.com/coordinate-s ... tml#160162
that's pretty comprehensive -- check it out.
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