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Re: ds-coordinates [#permalink]
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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Re: ds-coordinates [#permalink]
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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Re: ds-coordinates [#permalink]
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

Add them together, then
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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Re: ds-coordinates [#permalink]
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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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Re: ds-coordinates [#permalink]
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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