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In the rectangular coordinate system, are the points (a, b)

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In the rectangular coordinate system, are the points (a, b) [#permalink] New post 10 Dec 2007, 15:44
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A
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C
D
E

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In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)
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 [#permalink] New post 10 Dec 2007, 15:56
I think it's C.

1) NS because it only tells us that the distances are proportional, e.g. 2, 3 : 4, 6..

2) NS because a = 2, b = 3, c=1, d=5 satisfies the condition but the points are not equidistant.

1&2) Sufficient
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 [#permalink] New post 10 Dec 2007, 16:22
applecrisp wrote:
I think it's C.

1) NS because it only tells us that the distances are proportional, e.g. 2, 3 : 4, 6..

2) NS because a = 2, b = 3, c=1, d=5 satisfies the condition but the points are not equidistant.

1&2) Sufficient


How do you prove that 1 and 2 together are sufficient.
I am getting E
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 [#permalink] New post 11 Dec 2007, 12:21
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(C) too for me :)

Equidistant to 0 means:
sqrt(a^2 + b^2) = sqrt(c^2 + d^2)

In other words, the points A(a,b) and C(c,d) are both on a same circle, centered at (0,0).

Stat1
a/b = c/d = 1/s where s is a real number different from 0.

That implies:
o b = s*a
o d = s*c

Could we see an anology? :)... It's the equation of a line : y = s*x, passing by 0(0,0).

From here, we understand that, such points, can (sometimes) or cannot (most of time) be equidistant to 0.

o If a=-c and b=-d, we have the special case of equidistance. The line shares the points with a circle centered at 0(0,0)
o If a=c and d=d, both points are confused and so are equidistant to 0(0,0)
o All other cases are represented by the line y = s*x and do not create an equidistance (0 is out of the line ;))

INSUFF.

Stat2
o sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)
<=> |a| + |b| = |c| + |d|

We have 3 combinations:
o If |a| = |c|, then |b| = |d| that implies that A and B are in 4 possible points : the vertice of a square centered at 0(0,0)... (a,b), (-a,b), (a,-b) (-a,-b)
o If |a| = |d|, then |b| = |c| that implies that A and B are symetrical to the line y = x or on points rotated from 0(0,0) by k * 90° of these points (creating similar points in cadran 2, 3 and 4)... All of them are on a same circle centered at 0(0,0)
o All other cases : especially |b| = 0 and c!=0,d!=0... |a| = |c| + |d|

Concreate examples:
o If a=c=1 and b=d=0, then sqrt(a^2 + b^2) = sqrt(1+0)=1 and sqrt(c^2 + b^2) = sqrt(1+0) = 1.... equidistant to 0
o If a=2, b=0 and c=1 and d=1, then sqrt(a^2+b^2) = sqrt(2^2 + 0) = 2 and sqrt(c^2 + d^2) = sqrt(1 + 1) = sqrt(2).... not equidistant to 0

INSUFF.

Both 1 and 2
We have:
o b=s*a and d = s*c
o |a| + |b| = |c| + |d|

That implies:
o |a| + |s*a| = |c| + |s*c|
<=>(1+|s|)*|a| = (1+|s|)*|c|
<=> |a| = |c|

So,
o |a| + |b| = |c| + |d|
<=> |a| + |b| = |a| + |d|
<=> |b| = |d|

Bingo!... We are on the special case of:
o If |a| = |c|, then |b| = |d| that implies that A and B are in 4 possible points : the vertice of a square centered at 0(0,0)... (a,b), (-a,b), (a,-b) (-a,-b)

SUFF.
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 [#permalink] New post 11 Dec 2007, 13:07
Fig wrote:
(C) too for me :)

That implies:
o |a| + |s*a| = |c| + |s*c|
<=>(1+|s|)*|a| = (1+|s|)*|c|
<=> |a| = |c|
So,
o |a| + |b| = |c| + |d|
<=> |a| + |b| = |a| + |d|
<=> |b| = |d|

Bingo!... We are on the special case of:
o If |a| = |c|, then |b| = |d| that implies that A and B are in 4 possible points : the vertice of a square centered at 0(0,0)... (a,b), (-a,b), (a,-b) (-a,-b)

SUFF.


For some reason, I got stuck there. Thanks Fig. OA is C.
  [#permalink] 11 Dec 2007, 13:07
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