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Coordinate Problem!

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by Ozlemg » Mon Jul 11, 2011 12:26 am
In the rectangular coordinate system, are the points (a,b) and (c,d) equidistant from the origin?

(1) a/b =c/d

(2) sq root a^2+sq root b^2=sq root c^2+sq root d^2


I could not understand why (2) is Not SF? In (2), the formula is pyhtagoras theorem's formula. And according to this formula, if the square roots of 2 points are equal to square roots of other 2 points, the distances should be same!? So they should be equidistant from point (0,0)??

OA C
Last edited by Ozlemg on Mon Jul 11, 2011 12:42 am, edited 1 time in total.
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by Frankenstein » Mon Jul 11, 2011 12:36 am
Hi,
Please post the complete question. Are the points (a,b) and (c,d) ?
If so, for them to be equidistant from origin, sqrt(a^2+b^2) = sqrt(c^2 + d^2)
But statement(2) is not the same.
Example: consider (3,4) and (7,0)
From statement(2): 3+4 = 7+0. But are they equidistant from origin?
No, distance of (3,4) is 5 and (7,0) is 7.
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by Ozlemg » Mon Jul 11, 2011 12:49 am
Thank you...

But the point I could not get is, how come the pythagoes theorem does not work? Lets assume 2 triangles. When sum of the sq roots of 2 sides of these two triangles are equal (a^2+b^2=D , c^2+d^2 = D), then distances should be equal?
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by amit2k9 » Mon Jul 11, 2011 1:17 am
a a=b=c=d=1 each or a=b=1 each and c=d=2 each. thus not sufficient.

b |a|+ |b| = |c| + |d|

thus a=b=1 each and c=d=1 each or c=2 and d=0. not sufficient.

a+b

thus a=b=c=d= +|-1,2,3 and so on.
hence C it is.
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by GMATGuruNY » Mon Jul 11, 2011 3:48 am
Ozlemg wrote:Thank you...

But the point I could not get is, how come the pythagoes theorem does not work? Lets assume 2 triangles. When sum of the sq roots of 2 sides of these two triangles are equal (a^2+b^2=D , c^2+d^2 = D), then distances should be equal?
Distance from the origin:
a^2 + b^2 = D^2. (Note that the sum of the squares of a and b equals not the distance itself but the SQUARE of the distance.)
Thus, D = √(a^2 + b^2)

c^2 + d^2 = D^2
Thus, D = √(c^2 + d^2).

If the points are equidistant:
√(a^2 + b^2) = √(c^2 + d^2)

Statement 2 gives a different equation:
√(a^2) + √(b^2) =√(c^2) + √(d^2)

√(a^2 + b^2) is not the same as √(a^2) + √(b^2)
If a=1 and b=1:
√(1^2 + 1^2) = √2.
√(1^2) + √(1^2) = 1+1 = 2.
Not the same.
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by nitin9003 » Mon Jul 11, 2011 6:47 am
I am facing a problem to solve this question, can any1 please explain the solution i.e. why the answer is C.
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by sanabk » Mon Jul 11, 2011 10:04 am
Nitin,
Hope you follow this explanation.
If (a,b) and (c,d) are equidistant from origin, a^2+b^2=c^2+d^2.
1) ad=bc
2) sqrt(a^2)+sqrt(b^2)= sqrt(c^2)+sqrt(d^2)
=> a+b=c+d
=> b(a+b)=b(c+d) Here multiply both the sides with "b"
=> b(a+b)=bc+bd
=> b(a+b)=ad+bd Here substitute bc=ad from 1)
=> b(a+b)=d(a+b)
=> b=d
If b=d then from 1) a=c
Thus a^2+b^2=c^2+d^2 is true.

Best!!!
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by haqiqatprast » Tue Jul 12, 2011 7:44 pm
@sanabk I am afraid that your first implication is incorrect.

|a| + |b| = |c| + |d| does NOT imply a + b = c + d. For example let a = 1, b = -1 and c = d = 1

Consider the conditions:

(1) a/b = c/d
(2) |a| + |b| = |c| + |d|

I like to visualize the problem in the 2-D coordinate system. Then we can see that (1) ensures that both triangles are similar and (2) ensures that they are congruent. Hence the answer is C.
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