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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]  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.
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  02 Jun 2009, 06:44
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mbaMission wrote:
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]  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]  02 Jun 2009, 17:23
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 ?
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  02 Jun 2009, 20:44
4
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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]  03 Jun 2009, 06:24
1
KUDOS
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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Re: In the rectangular coordinate system, are the points (a, [#permalink]  08 Jun 2009, 22:29
GMAT TIGER wrote:
mbaMission wrote:
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.

why from a^2 + b^2= c^2 + d^2 => a+b=c+d ?
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  19 Aug 2009, 18:29
ngoctraiden1905 wrote:
why from a^2 + b^2= c^2 + d^2 => a+b=c+d ?

That is not so.

What other posters said is that from (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) => a+b=c+d

That is actually incorrect. Should be |a|+|b|=|c|+|d|.
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  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]  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]  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.

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Re: In the rectangular coordinate system, are the points (a, [#permalink]  05 Feb 2011, 13:27
1
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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.

your explanation is great as always ! thaaanks
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  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

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

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]  05 Feb 2011, 19:36
Bunuel is right! Thanks for the explanation man....
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  19 Feb 2011, 18:12
Entwistle 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

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

I liked your explanation, of course Bunuel was as helpful as ever too.
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  21 Dec 2011, 09:34
1
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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

HTH
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  27 Jul 2012, 20:49
Hi there, i was just wondering if the way i do it is correct!

Statement 1: insuff
Statement 2 : |a|+|b|=|c|+|d|

Our goal is to prove that a^2 + b^2 = c^2 + d^2

(1+2)

Square both sides in stmt 2.

We have a^2 + b^2 + 2|a||b| = c^2 + d^2 + 2|c||d| ----------- *
From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2|a||b| from LHS and 2|c||d| from equation *.

Please tell me that i am correct! =)

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Re: In the rectangular coordinate system, are the points (a, [#permalink]  08 Aug 2012, 02:22
1
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reagan wrote:
Hi there, i was just wondering if the way i do it is correct!

Statement 1: insuff
Statement 2 : |a|+|b|=|c|+|d|

Our goal is to prove that a^2 + b^2 = c^2 + d^2

(1+2)

Square both sides in stmt 2.

We have a^2 + b^2 + 2|a||b| = c^2 + d^2 + 2|c||d| ----------- *
From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2|a||b| from LHS and 2|c||d| from equation *.

Please tell me that i am correct! =)

Reagan

Absolutely, I found this approach much better. Infact I am wondering why we are targetting absolute values in the equation.

(a+b)^2 = a^2+b^2+2ab [no absolute |a|, |b| needed]

From
1) we know ab = cd
2) we know a + b = c + d. and hence (a+b)^2 = (c+d)^2

Combining 1) & 2) we can very well see that a^2+b^2 = c^2+d^2
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  01 Nov 2013, 03:40
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Re: In the rectangular coordinate system, are the points (a, [#permalink]  25 May 2014, 07:13
Awesome discussion and solution analysis.

Thank you all!
Re: In the rectangular coordinate system, are the points (a,   [#permalink] 25 May 2014, 07:13

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