Co-ordinate Geomatry

[akpareek]

In the xy plane, are the points (r,s) and (u,v) equidistance from the origin ?

1. r+s=1
2. u=1-r and v=1-s

[kvcpk]

In the xy plane, are the points (r,s) and (u,v) equidistance from the origin ?

1. r+s=1
2. u=1-r and v=1-s

Distance of (r,s) from origin = r^2 + s^2
Question is asking IS r^2+s^2 = U^2 + v^2
or u^2 - r^2 = s^2-v^2

St1 doesnt give info about (u,v). INSUFF

St2:
u=1-r and v=1-s
u^2 = 1+r^2 -2r
u^2-r^2 = 1-2r
Similarly, v^2 -s^2 = 1-2s
No info available yet. because we do not know about r and s values.
INSUFF

Combining:
u^2-r^2 = 1-2r
v^2 -s^2 = 1-2s
Add these two
u^2+v^2-(r^2+s^2) = 2-2(r+s)
u^2+v^2-(r^2+s^2) = 2-2(1)
u^2+v^2-(r^2+s^2) = 0
Hence
u^2+v^2=(r^2+s^2)
SUFF

pick C
r+s =1

[clock60]

one more way to solve
is r^2+s^2=u^2+s^2, or
(r+s)^2-2rs=(u+v)^2-2uv

clearly neither 1 st nor the second sts are insufficient to answer. but together
u+r=1
v+s=1
let us add equations
u+v+(r+s)=2, and from 1 st r+s=1, so u+v=1 also (u+v+1=2, u+v=1)

so we proved that for equation
(r+s)^2-2rs=(u+v)^2-2uv. r+s=u+v=1, and we can cancel both and left with
is rs=uv, if we manage to prove that rs=uv so our points are equidistant

r=1-u, v=1-s
(1-u)*s=u(1-s), s-su=u-su. cancel us , so s=u, and r=v
so C