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# DS-Equidistant from Origin.jpg

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Manager
Joined: 21 May 2008
Posts: 85

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DS-Equidistant from Origin.jpg [#permalink]

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10 Oct 2008, 06:43
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VP
Joined: 30 Jun 2008
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Re: DS-Equidistant from origin [#permalink]

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10 Oct 2008, 08:07
In the rectangular co-ordinate system, are the points (r,s) and (u,v) equidistant from the original ?

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

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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Re: DS-Equidistant from origin   [#permalink] 10 Oct 2008, 08:07
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# DS-Equidistant from Origin.jpg

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