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# ds-coordinates

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18 Oct 2008, 20:52
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In a rectangular coordinate system ,are the points (r,s) and (u,v)
equidistant from origin?
1)r+s=1
2)u=1-r and v=1-s

Kindly HELP solving this since im not able to get to the answer !!!
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Joined: 30 Jun 2008
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18 Oct 2008, 20:59
spriya wrote:
In a rectangular coordinate system ,are the points (r,s) and (u,v)
equidistant from origin?
1)r+s=1
2)u=1-r and v=1-s

Kindly HELP solving this since im not able to get to the answer !!!

Origin is (0,0). Distance between origin and (r,s) is SQRT(r² + s²) and Distance between origin and (u,v) is SQRT(u² + v²)

so we have to prove if SQRT(u² + v²) = SQRT(r² + s²)

Now (1) is insufficient since it provides us NO information reg u and v

(2) u = 1-r and v=1-s

distance from (0,0) and (u,v) is = SQRT(u² + v²)
= SQRT((1-r)² + (1-s)²)
= SQRT(1 + r²- 2r + 1 + s² - 2s)
= SQRT(2 -2(r+s) + r² + s²)

Now we are stuck here. 2 also insufficient.

Combining 1 and 2 we get -- SQRT(u² + v²) = SQRT(2 -2(r+s) + r² + s²)
from 1 we get r+s =1 on substituting this in the above equation we get SQRT(u² + v²) = SQRT(r² + s²)

Hence the answer must be C
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18 Oct 2008, 21:25
amitdgr wrote:
spriya wrote:
In a rectangular coordinate system ,are the points (r,s) and (u,v)
equidistant from origin?
1)r+s=1
2)u=1-r and v=1-s

Kindly HELP solving this since im not able to get to the answer !!!

Origin is (0,0). Distance between origin and (r,s) is SQRT(r² + s²) and Distance between origin and (u,v) is SQRT(u² + v²)

so we have to prove if SQRT(u² + v²) = SQRT(r² + s²)

Now (1) is insufficient since it provides us NO information reg u and v

(2) u = 1-r and v=1-s

distance from (0,0) and (u,v) is = SQRT(u² + v²)
= SQRT((1-r)² + (1-s)²)
= SQRT(1 + r²- 2r + 1 + s² - 2s)
= SQRT(2 -2(r+s) + r² + s²)

Now we are stuck here. 2 also insufficient.

Combining 1 and 2 we get -- SQRT(u² + v²) = SQRT(2 -2(r+s) + r² + s²)
from 1 we get r+s =1 on substituting this in the above equation we get SQRT(u² + v²) = SQRT(r² + s²)

Hence the answer must be C

Yes OA is C !!i just did not want to do so much of calculation !!so got this one wrong !!just guessed !!
Is there any simpler way to do this ,like which takes less time !!
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21 Oct 2008, 02:49
There is a simpler way but its similar.

Now we need to prove that u^2 + v^2 = (1-r)^2 + (1-s)^2

RHS = 1-2r-2s + u^2 + s^2

Now we can only come to our desired result if (1-2r -2s) =0 or in other words r+s =0. This condition is given by A. So we have to combine A with B for our desired result. Hence C
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Thanks
rampuria

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21 Oct 2008, 02:55
rampuria wrote:
There is a simpler way but its similar.

Now we need to prove that u^2 + v^2 = (1-r)^2 + (1-s)^2

RHS = 1-2r-2s + u^2 + s^2

Now we can only come to our desired result if (1-2r -2s) =0 or in other words r+s =0. This condition is given by A. So we have to combine A with B for our desired result. Hence C

I don't get the portion in red
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21 Oct 2008, 10:59
try this:

1)r+s=1 NOT SUFFICIENT

2)u=1-r and v=1-s NOT SUFFICIENT

If using 1) + 2)

u=1-r
Can be written as: r= 1-u

v=1-s
Can be written as: s= 1-v

r+s = 2 - u - v
From 1) r+s= 1

1 = 2 - u - v
u+v =1

Both 1), 2) are giving the same value (both =1), thus, equidistant.

What do you guys think?
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21 Oct 2008, 12:26
1
KUDOS
spriya wrote:
In a rectangular coordinate system ,are the points (r,s) and (u,v)
equidistant from origin?
1)r+s=1
2)u=1-r and v=1-s

Kindly HELP solving this since im not able to get to the answer !!!

Here is simple way.

1)
r+s=1
2)
u=1-r --> u+r=1
and v=1-s --> v+s=1

r+s=1=u+r --> s=u
r+s=1=v+s --> r=v

So (r,s) and(u,v) are same points.. so must be equidistant from origin.
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21 Oct 2008, 20:07
x2suresh wrote:
spriya wrote:
In a rectangular coordinate system ,are the points (r,s) and (u,v)
equidistant from origin?
1)r+s=1
2)u=1-r and v=1-s

Kindly HELP solving this since im not able to get to the answer !!!

Here is simple way.

1)
r+s=1
2)
u=1-r --> u+r=1
and v=1-s --> v+s=1

r+s=1=u+r --> s=u
r+s=1=v+s --> r=v

So (r,s) and(u,v) are same points.. so must be equidistant from origin.

thats a really neat way of doing this problem
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Re: ds-coordinates   [#permalink] 21 Oct 2008, 20:07
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