I agree with u Amitava IMO also it should be E

(1) r + s = 1
no detail abt u,v INSUFF

(2) u = 1 – r and v = 1 – s

here u,v can be anthing based upon r & s INSUFF (same or diff)

Combine: now (r+s)^2 = r^2 +s^2 + 2rs = 1
(to be equidsiatnt from origin the pts must lie on the same circle )

so r^2 + s^2 = 1-2rs

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

so (u+v)^2 = 1

so u^2 +v^2 = 1- 2uv

so u^2 + v^2 = 1-2(1-r)(1-s)
so if 1-r =r & 1-s =s i.e both are 0.5 & pts are same then they will be equidistant otherwise not.

E

I beileve OA can be C if the q said both pts lie in the first quadrant.

Hi Samir,

This is what I did and got the ans as C. Can you pls tell me what you think

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

So combined we know

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

Since r + s = 1 and u+v = 1

r+s = u+v

So they should be equidastant right?

r,s = 10, -9 u,v = 40, -39 are not equidistant

even 0.5,0.5 , .25,.75 are not equidistant

0.5,.5 & .5,.5 is equidistant both can be same pts

Thanks a lot Samir.

The answer MUST be C.

For (r,s) and (u,v) to be equidistant from Origin, they have to satisfy
r^2 + s^2 = u^2 + v^2

And this is only arrived at using both (1) and (2). Hence C.

Calista.