GMAT PREP SOFTWARE - CAT 2, #6: Data Sufficiency
In a rectangular coordinate system, are points (r, s) and (u, v) equidistant from the origin?
1. r+s = 1
2. u=1-r and v=1-s
Answer is C (Together they are sufficient). Can you please tell me how to answer this question?
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- guest612
- sheetal
If two points are equidistant, then their distance from origin is the same.
Distance of (r,s) from origin = (r^2) + (s^2)
Distance of (u,v) from origin = (u^2) + (v^2).
So the question is asking is (u^2) + (v^2) = (r^2) + (s^2) ?
(1) and (2) are both insufficient, so (a), (d) and (b) are out.
Applying squares to (2), gives
(u^2) = 1 + (r^2) - 2r
(v^2) = 1 + (s^2) - 2s
---------------------------------- => Adding the above equations give the below result
(u^2) + (v^2) = (r^2) + (s^2) + 2 (1 - r - s) ... (3)
From (1) we know that
r + s = 1
=> 1 - r - s = 0. ..(4)
Substituting eqn (4) in Eqn (3) gives us
(u^2) + (v^2) = (r^2) + (s^2) + 2 *0
=>(u^2) + (v^2) = (r^2) + (s^2) , so the ans is TRUE
C is the answer.
Distance of (r,s) from origin = (r^2) + (s^2)
Distance of (u,v) from origin = (u^2) + (v^2).
So the question is asking is (u^2) + (v^2) = (r^2) + (s^2) ?
(1) and (2) are both insufficient, so (a), (d) and (b) are out.
Applying squares to (2), gives
(u^2) = 1 + (r^2) - 2r
(v^2) = 1 + (s^2) - 2s
---------------------------------- => Adding the above equations give the below result
(u^2) + (v^2) = (r^2) + (s^2) + 2 (1 - r - s) ... (3)
From (1) we know that
r + s = 1
=> 1 - r - s = 0. ..(4)
Substituting eqn (4) in Eqn (3) gives us
(u^2) + (v^2) = (r^2) + (s^2) + 2 *0
=>(u^2) + (v^2) = (r^2) + (s^2) , so the ans is TRUE
C is the answer.
- guest612
wow
that was great! thanks very much.
quick question, how does squaring the coordinates demonstrate it is equidistant? I think I've got it but I want to confirm. Is it because it takes into account the absolute value of the coordinates by squaring it? And as on a number line, we are looking for the distance on the coordinate plane and squaring the coordinates eliminates the discrepancy of the negative values and produces the value of the distance?
quick question, how does squaring the coordinates demonstrate it is equidistant? I think I've got it but I want to confirm. Is it because it takes into account the absolute value of the coordinates by squaring it? And as on a number line, we are looking for the distance on the coordinate plane and squaring the coordinates eliminates the discrepancy of the negative values and produces the value of the distance?
- sheetal
For calculating the distance between two points, we use the Pythagorean theorem. The length of the diagonal gives the distance between two points.
For example if the two points are (1,3) and (7,-5), then horizontal distance = (7-1) = 6
The vertical distance = (-5-3) = -8. The negative sign doesn't matter as the change in y coordinate = 8.
The length of the diagonal will give the distance between the two points.
Use Pythagorean theorem to calculate the length (L):
(L^2) = (6^2) + (8^2)
=> L = 10.
Similarly if one of the points lies on the centre (0,0), distance between lets say (0,0) and (3,-4) will be
(L^2) = (3^2) + (4^2)
=> L = 5.
For example if the two points are (1,3) and (7,-5), then horizontal distance = (7-1) = 6
The vertical distance = (-5-3) = -8. The negative sign doesn't matter as the change in y coordinate = 8.
The length of the diagonal will give the distance between the two points.
Use Pythagorean theorem to calculate the length (L):
(L^2) = (6^2) + (8^2)
=> L = 10.
Similarly if one of the points lies on the centre (0,0), distance between lets say (0,0) and (3,-4) will be
(L^2) = (3^2) + (4^2)
=> L = 5.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
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5 posts Page 1 of 1