From (1), the two coordinate pairs may be equidistant as 2/3 = 2/3 however may also not be equidistant as 2/3 = 10/15. So, (1) is insufficient.
Considering (2): sqrt (a^2) means |a|. So, the absolute value or magnitudes of a + b = c + d. So, the two coordinate pairs may be equidistant as 2 + 2 = 2 + 2 however may also not be equidistant as 0 + 4 = 2 + 2 (here, the right hand side would yield a distance of 2root2 from origin).
(1 + 2):
The second cases in the analysis for each statement above are now removed. That is 2/3 now cannot equal 10/15 as this would violate (2). And 0 + 4 now cannot equal 2 + 2 as this would violate (1). Possible cases for (a,b) and (c,d) respectively are: (1,2) and (-1, -2); or (-1, 2) and (1,-2), etc. In any case, the two coordinate pairs will be equidistant from origin.
MGMAT: Coordinate Geom - Need explanation
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If (a,b) and (c,d) have to be equidistant from the origin, a^2 + b^2 = c^2 + d^2.
Consider first (1) alone.
Let a = 2, b = 3, c = 4 and d = 6.
Here a/b = c/d = 2/3, but a^2 + b^2 = 13 and c^2 + b^2 = 52.
So (a,b) and (c,d) are not equidistant from the centre.
Next let a = 2, b = 3, c = -2 and d = -3.
Here a/b = c/d = 2/3 and also a^2 + b^2 = c^2 + d^2 = 13.
So (a,b) and (c,d) are equidistant from the centre.
Since nothing definite can be said, (1) alone is not sufficient.
Next consider (2) alone.
It means lal+lbl =lcl+ldl.
Let a = 2, b = 8, c = 5 and d = 5.
Here lal+lbl =lcl+ldl = 10, but a^2 + b^2 = 68 and c^2 + d^2 = 50.
So (a,b) and (c,d) are not equidistant from the centre.
Next let a = 2, b = 8, c = 8 and d = 2.
Here lal+lbl = lcl+ldl = 10 and a^2 + b^2 = c^2 + d^2 = 68.
So (a,b) and (c,d) are equidistant from the centre.
Again nothing definite can be said and so (2) alone is not sufficient.
Next combine both the statements together and check.
Let a/b = c/d = k.
So a = kb and c = kd.
Also from (2), lkbl+lbl = lkdl+ldl.
So (lkl + 1)(lbl - ldl) =0.
Hence either lbl = ldl or lkl = -1 (not possible).
So lbl = ldl, and lal = lkbl =lkdl = lcl.
In this case a^2 + b^2 = c^2 + d^2.
So we can say that both points are equidistant from the centre.
The correct answer is (C).
Consider first (1) alone.
Let a = 2, b = 3, c = 4 and d = 6.
Here a/b = c/d = 2/3, but a^2 + b^2 = 13 and c^2 + b^2 = 52.
So (a,b) and (c,d) are not equidistant from the centre.
Next let a = 2, b = 3, c = -2 and d = -3.
Here a/b = c/d = 2/3 and also a^2 + b^2 = c^2 + d^2 = 13.
So (a,b) and (c,d) are equidistant from the centre.
Since nothing definite can be said, (1) alone is not sufficient.
Next consider (2) alone.
It means lal+lbl =lcl+ldl.
Let a = 2, b = 8, c = 5 and d = 5.
Here lal+lbl =lcl+ldl = 10, but a^2 + b^2 = 68 and c^2 + d^2 = 50.
So (a,b) and (c,d) are not equidistant from the centre.
Next let a = 2, b = 8, c = 8 and d = 2.
Here lal+lbl = lcl+ldl = 10 and a^2 + b^2 = c^2 + d^2 = 68.
So (a,b) and (c,d) are equidistant from the centre.
Again nothing definite can be said and so (2) alone is not sufficient.
Next combine both the statements together and check.
Let a/b = c/d = k.
So a = kb and c = kd.
Also from (2), lkbl+lbl = lkdl+ldl.
So (lkl + 1)(lbl - ldl) =0.
Hence either lbl = ldl or lkl = -1 (not possible).
So lbl = ldl, and lal = lkbl =lkdl = lcl.
In this case a^2 + b^2 = c^2 + d^2.
So we can say that both points are equidistant from the centre.
The correct answer is (C).
Rahul Lakhani
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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
