f2001290 wrote:18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 – r and v = 1 – s

Please Explain

… if we join the origin to the points (r,s) and (u,v) .. then we have two triangles both of them wuld be right angle triangles … and the hypotenuse of the triangles wuld be the distance from the center … and the sides for the 2 triangle wuld be r, s for the first one and u,v for the second one ..

so what the q basically asks is if r^2+s^2 = u^2+v^2 …

.. the first statement says that r+s=1 .. which does not really help .. so insufficient ..

… from the second statement we have
u^2= 1+r^2-2r
v^2=1+s^2-2s …
adding both of them we have u^2+v^2=2+r^2+s^2-2(r+s) .. so we have
(u^2+v^2) - (r^2+s^2) = 2 – 2(r+s) now remember if the 2 points were equidistant then (u^2+v^2) – (r^2+s^2) wuld be equal to 0 .. but for it to be 0 (r+s) has to be equal to 1 but we don’t have anything in the second statement that says so .. so second statement is also insufficient

now combine both the statement and from the first statement we have r+s=1 … and by substituting this in the eqn (u^2+v^2)-(r^2+s^2)=2-2(r+s) .. we get (u^2+v^2)-(r^2+s^2)=0 .. and that means the two points are equidistant from the origin .. so the answer is C .. post if any doubts ..

gabriel wrote:

f2001290 wrote:18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 – r and v = 1 – s

Please Explain

… if we join the origin to the points (r,s) and (u,v) .. then we have two triangles both of them wuld be right angle triangles … and the hypotenuse of the triangles wuld be the distance from the center … and the sides for the 2 triangle wuld be r, s for the first one and u,v for the second one ..

so what the q basically asks is if r^2+s^2 = u^2+v^2 …

.. the first statement says that r+s=1 .. which does not really help .. so insufficient ..

… from the second statement we have
u^2= 1+r^2-2r
v^2=1+s^2-2s …
adding both of them we have u^2+v^2=2+r^2+s^2-2(r+s) .. so we have
(u^2+v^2) - (r^2+s^2) = 2 – 2(r+s) now remember if the 2 points were equidistant then (u^2+v^2) – (r^2+s^2) wuld be equal to 0 .. but for it to be 0 (r+s) has to be equal to 1 but we don’t have anything in the second statement that says so .. so second statement is also insufficient

now combine both the statement and from the first statement we have r+s=1 … and by substituting this in the eqn (u^2+v^2)-(r^2+s^2)=2-2(r+s) .. we get (u^2+v^2)-(r^2+s^2)=0 .. and that means the two points are equidistant from the origin .. so the answer is C .. post if any doubts ..

Agree with C. - Excellent post gabriel.

an excellent explanation. but sometime's its faster to try and reason it out with a random set of values and see if the conditions are satisified for example let (r,s)=(4,3) and (u,v)=(3,4) C gets satisfied. But dont depend on only one value take more sets

abkhan wrote:an excellent explanation. but sometime's its faster to try and reason it out with a random set of values and see if the conditions are satisified for example let (r,s)=(4,3) and (u,v)=(3,4) C gets satisfied. But dont depend on only one value take more sets

... hmmm .. i dont know abt that .. first of the vlues u have chosen do not match the condiiton given .. second of ... somehow i am not in favour of using ' random values" for a DS .. especially tricky ones ... nywayz watever works for u ..

thanks a lot gabriel...great explanation

gabriel wrote:

f2001290 wrote:18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 � r and v = 1 � s

Please Explain

� if we join the origin to the points (r,s) and (u,v) .. then we have two triangles both of them wuld be right angle triangles � and the hypotenuse of the triangles wuld be the distance from the center � and the sides for the 2 triangle wuld be r, s for the first one and u,v for the second one ..

.

Can anyone explain how 2 triangles are formed??

Instead of using the two triangle thing which is a little confusing

Think about the distance formula. If 2 points are equidistant , obviously the distance is the same

D = sqrt ( (x-x1)^2 - (y-y1)^2 )

When one point is origin, hence
d = sqrt (x^2 - y^2)

And which gives
r^2 + s^2 = u^2 + v^2

18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 - r and v = 1 - s

If I such substitute (1) is each of the 2 eq. of (2)
u = 1-r but from (1) r+s = 1 so
u = r+s-r = s
v = 1-s = r+s-s = r
so (r,s) and (u,v) are same points
somethings funny

The points are actually on a circle with origin (0,0)