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

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In the rectangular coordinate system, are the points (a, [#permalink] New post 02 Jun 2009, 03:05
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In the rectangular coordinate system, are the points \((a, b)\) and \((c, d)\) equidistant from the origin?

(1) \(\frac{a}{b} =\frac{c}{d}\)

(2) \(\sqrt{a^2} + \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Apr 2015, 03:31, edited 1 time in total.
Added the OA.
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 02 Jun 2009, 06:44
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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) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2)


(1) If a/b = c/d, a and b could be 100 and c and d could be 1 or a, b, c and d could be 1. NSF.

(2) If (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2), then a+b = c+d. In this case, if a = 9 and b = 1, and c =d=5, a+b = c+d is true. but they have different distance.NSF.

From 1 and 2, a must be equal to c and b must be equal to d. So Suff.

Thats C.
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 02 Jun 2009, 11:13
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

This singnifies that the sign combination on both sides of = is the same.
i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well.
However since a,b / c,d can take any value , INSUFFICIENT.

(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2)
Since we do not know anything about signs of a,b / c,d - INSUFFICIENT.


Using 1 and 2 together ,
using (1) the second statement drills down to
a+b=c+d (the sign combination on both sides of = is the same.)

Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point.
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 02 Jun 2009, 20:44
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goldeneagle94 wrote:
amolsk11 wrote:
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

This singnifies that the sign combination on both sides of = is the same.
i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well.
However since a,b / c,d can take any value , INSUFFICIENT.

(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2)
Since we do not know anything about signs of a,b / c,d - INSUFFICIENT.


Using 1 and 2 together ,
using (1) the second statement drills down to
a+b=c+d (the sign combination on both sides of = is the same.)

Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point.


Is there a way we can derive, using these two equations, that the two points are equidistant ?



For the derivation part:

a^2 + b^2 = c^2 + d^2

b^2[(a^2/b^2) + 1] = d^2[(c^2/d^2 +1)] --------> 1

from I

a/b = c/d
squaring both sides
a^2/b^2 = c^2/d^2
putting above in 1 and simplifying

b^2[(c^2/d^2) + 1] = d^2[(c^2/d^2 +1)]
simplifying

b^2 = d^2
or |b| = |d|
=> |a| = |c|

therefore points are equidistant from (0,0) or Origin.
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 03 Jun 2009, 06:24
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in order for the points to be equidistant

a2 + b2 = c2+ d2 -----> eq X

1--> insufficent (has already been discussed)
2--> insufficent (has already been discussed)

combining both,
equation 2 can be written as a+b = c+d

eq1 a/b = c/d
a+b/b = c+d/d, from eq 1 from b=d ----3

similarly, b/a = d/c, a+b/a = d+c/c from eq 1 a = c ----4


plug in values of a, b in eq X... it satifies.

Hence points are equidistant. Answer is C
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In the rectangular coordinate system, are the points (a, b) [#permalink] New post 10 Apr 2010, 22:25
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In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) \(\frac{a}{b}=\frac{c}{d}\)

(2) \(\sqrt{a^2}+ \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)

Last edited by Bunuel on 07 Apr 2012, 07:24, edited 1 time in total.
Edited the question and added the OA
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Re: ManhttanGMAT Practice CAT [#permalink] New post 11 Apr 2010, 00:11
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nsp007 wrote:
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) \(\sqrt{a^2}+ \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)

Will post OA later.



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

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Answer: C.
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Re: ManhttanGMAT Practice CAT [#permalink] New post 08 Sep 2010, 01:22
Bunuel, had condition (2) simply said a+b=c+d (instead of the squares and square root) how would have the answer changed?

Also in the current question - is a^2+b^2=c^2+d^2?
When I square both sides of (2), I get a^2+b^2+2sqrt(a^2b^2)=c^2+d^2+2sqrt(c^2d^2)
so if ab=cd then this is satisfied. however (1) only gives me ad=bc, how do I infer ab=cd from that? I am following a different process, but I should end up with the same answer. Not sure where am I wrong?
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Re: ManhttanGMAT Practice CAT [#permalink] New post 09 Sep 2010, 00:42
I have a question here bunuel. How did you get a = c * x and b= d*x??

Because from this it means that x = b/d = a/c

Can you please explain??
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Re: ManhttanGMAT Practice CAT [#permalink] New post 09 Sep 2010, 00:55
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amitjash wrote:
I have a question here bunuel. How did you get a = c * x and b= d*x??

Because from this it means that x = b/d = a/c

Can you please explain??


It's the same: \(\frac{b}{d} = \frac{a}{c}\) --> \(bc=ad\)--> \(\frac{c}{d}=\frac{a}{b}\).

Given: \(\frac{a}{b}=\frac{c}{d}=\frac{cx}{dx}\) (as the ratios are equal then there exist some \(x\) for which \(a=cx\) and \(b=dx\)).
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Re: points equidistant from origin? [#permalink] New post 16 Oct 2010, 14:52
(1) knowing these proportions does not help me solve it, because for example if 3/1 = 9/3 , point (a,b) will be closer to the origin than point (b,c)

(2) this statement tells that |a|+|b|=|c|+|d| , which is still not sufficient because we lack information about the correlation between |a| and |b|,and |c|and|d|.

But if we combine the two statements together we will have this correlation from statement (1) and then both statements taken together will be sufficient.

Answer should be C.
What is the OG answer?
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Re: points equidistant from origin? [#permalink] New post 16 Oct 2010, 15:21
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Orange08 wrote:
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

a. \(a/b = c/d\)

b. \(\sqrt{a^2}+\sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)


Distance of \((x,y)\) from origin is \(\sqrt{x^2+y^2}\)
So we need to answer \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\) ?

(1) a/b=c/d ... doesnt really help in proving or disproving. Insufficient

(2) This is equivalent to saying \(|a|+|b|=|c|+|d|\). Again insufficient to say anything about the statement we have.

(1+2) a/c=b/d=x say (needs, c,d to be non zero)

a = cx
b = dx

|a|+|b|=|c|+|d|
|cx| - |c| = |d| - |dx|
|c|(|x|-1)=|d|(1-|x|)
(|c|+|d|)(|x|-1)=0
Since c,d are non-zero means |x|=1
So either a=c & b=d OR a=-c or b=-d
Either case a^2=c^2 and b^2=d^2
Hence \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)
Sufficient

Answer is (c)
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Re: Difficult Geometry DS Problem [#permalink] New post 12 Jan 2011, 18:43
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abmyers wrote:
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) \(\sqrt{a^2}+\sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


You can take values to solve this question quickly:

Statement 1: a/b = c/d
(a,b) and (c,d) may be equidistant from the origin e.g. (1, 1) and (-1, -1) or they may not be e.g. (1, 1) and (2, 2). Not sufficient.

Statement 2: \(\sqrt{a^2}+\sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)
That is, |a|+|b|=|c|+|d|
(a,b) and (c,d) may be equidistant from the origin e.g. (1, 3) and (-1, -3) or they may not be e.g. (1, 3) and (2, 2)

Using both together, |a|+|b|=|c|+|d| and a/b = c/d.
This means that if a/c = 1/2, c/d cannot be 2/4 or -3/-6 etc. c/d has to be either 1/2 or (-1)/(-2). Similarly, if a/b = (-1)/2, c/d = (-1)/2 or 1/(-2))
Any pair of such points will be equidistant. Answer (C).
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 05 Feb 2011, 11:45
Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 05 Feb 2011, 12:06
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tinki wrote:
Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses

a/b=c/d <---- Equation 1

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

Add 1 to both sides of Equation 1

(a+b)/b=(c+d)/d

Now a+b=c+d => 1/b=1/d

Thus, b=d
Similarly,
b/a=d/c
Adding 1 to both sides of above equation:
(b+a)/a=(d+c)/c
a+b=d+c => 1/a=1/c
Therefore a=c
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 05 Feb 2011, 12:53
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tinki wrote:
Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses


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

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Answer: C.
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 05 Feb 2011, 13:27
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Bunuel wrote:
tinki wrote:
Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses


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

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Answer: C.


your explanation is great as always ! thaaanks
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 05 Feb 2011, 13:32
Expert's post
AmrithS wrote:
tinki wrote:
Official explanation says:
"If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

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a/b=c/d <---- Equation 1

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

Add 1 to both sides of Equation 1

(a+b)/b=(c+d)/d

Now a+b=c+d => 1/b=1/d

Thus, b=d
Similarly,
b/a=d/c
Adding 1 to both sides of above equation:
(b+a)/a=(d+c)/c
a+b=d+c => 1/a=1/c
Therefore a=c


No that's not correct.

|a|+|b|=|c|+|d| doesn't mean that a+b=c+d (consider a=b=1 and c=d=-1). So from |a|+|b|=|c|+|d| and a/b=c/d we can not derive that a=c and b=d. What we can derive is that |a|=|c| and |b|=|d|. Refer to my post above for complete solution.
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Re: In the rectangular coordinate system, are the points (a, [#permalink] New post 21 Dec 2011, 09:34
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Here's a simpler solution......

We need to prove a^2 + d^2 = c^2 + b^2

Simpler way is to rearrange statement 2 viz. |a| + |b| = |c| + |d| as
1. |a| - |d| = |c| - |b| and then square both sides. We get -
2. a^2 + d^2 - 2*|a|*|d| = c^2 + b^2 - 2*|c|*|b|.

Since ad = bc as per statement 1,

3. |ad| = |bc| => |a|.|d| = |b|.|c| (Rule -> Abs of Product = Product of Abs)

So we can cancel the third term out from both sides of equation 2. to get the desired equation

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Re: ManhttanGMAT Practice CAT [#permalink] New post 27 May 2012, 22:56
Bunuel wrote:
nsp007 wrote:
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) \(\sqrt{a^2}+ \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)

Will post OA later.



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

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Answer: C.


I am getting a different final result for answer (C). Here is my approach:
\(\sqrt{a^2} - \sqrt{d^2} = \sqrt{c^2} - \sqrt{b^2}\) ----from statement (2)

Squaring both sides

\(a^2 + d^2 - 2\sqrt{a^2*d^2} = c^2 + b^2 - 2\sqrt{b^2*c^2}\)

from statement (1) we know that ad=bc --> a^2*d^2 = b^2*c^2

Cancelling last term of both sides, we get

a^2 + d^2 = c^2 + b^2
Thus, a^2 - b^2 = c^2 - d^2

Not able to figure out where I went wrong. Please suggest!

Thanks
Re: ManhttanGMAT Practice CAT   [#permalink] 27 May 2012, 22:56

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