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

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Director
Joined: 03 Jul 2003
Posts: 654
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In the rectangular coordinate system, are the points (r, s) [#permalink]  13 Dec 2003, 20:59
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In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 тАУ r and v = 1 тАУ s
Director
Joined: 03 Jul 2003
Posts: 654
Followers: 2

Kudos [?]: 32 [0], given: 0

Director
Joined: 14 Oct 2003
Posts: 588
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 10 [0], given: 0

okay. we want to know whether pts (r,s) and (u,v) are equidistant from the origin (point 0,0) in the coordinate plane.

that means you can have values for r,s and u,v as (-1,0) and (1,0) or (-3,3) and (3,3) or (0,4) and (0,-4) etc etc. you get the picture

1. does r+s=1 tell us anything about point (u,v)? insufficient!

2. rewrite this to u+r=1 and v+s=1

okay few scenarios here

if (r,s) = (0,1) and (u,v) = (1,0) the two coordinates would be equidistant

however if you had values for (r,s) = (-2,-3) and (u,v) = (3,4) then they would NOT be equidistant.
Insufficient

However, they would be sufficient if you had 1 and 2.
no matter how what scenario you run through for both coordinates, with both 1 and 2 you'll have equidistant points.

try it out:

(r,s) = .5, .5 then (u,v) must equal .5 and .5
(r,s) = 0,1 then (u,v) must equal 1,0
(r,s) = 1,0 then (u,v) must equal 0,1
(r,s) = -1,2 then (u,v) must equal 2,-1

hope this helps
Director
Joined: 13 Nov 2003
Posts: 971
Location: Florida
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agree
when r+s = 1, then r,u and v,s will always have the opposite co-ordinates
SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

Kudos [?]: 43 [0], given: 0

C is the answer. See the following explaination

let x1 be the distance from (0,0) to (r,s)
so
sqr(x1) = sqr(r) + sqr(s)

similarly x2 be the distance from (0,0) to (u,v)
so
sqr(x2) = sqr(u) + sqr(v)

Condition 1 does not say anything aout relation between (r,s) and (u,v)
so it cannot convey anything

Condition 2 however talks about relation between (r,s) and (u,v)

sqr(x2) = sqr(u) + sqr(v) = sqr(1-r) + sqr(1-s)

= 1 + sqr(r) - 2r + 1 + sqr(s) - 2s

= 1 + sqr(r) + sqr(s) - 2( r+s )
Again we cannot understand anything from this condition

-------------------------------------------------------------

if we combine condition 1 and 2
u + v = 2-(r+s) = 1 ( since r+s=1 from condition 1 )
so u+v = r+s

We can say sqr(u) + sqr(v) = sqr((u+v) - 2uv = 1-2uv ( which sqr of x2 )

Now we dont know the value of uv so we use second condition to
multiply u and v
uv = (1-r) * (1-s) = 1+rs - (r+s) = 1 + rs -1 = rs
therefore uv = rs

We can say sqr(r) + sqr(s) = sqr(r+s) - 2rv = 1-2sr = 1-2uv

Ultimatey x1 = x2 hence they are equidistant
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