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Re: In the rectangular coordinate system, are the points (a, [#permalink]
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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]
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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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Re: ManhttanGMAT Practice CAT [#permalink]
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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]
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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]
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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: In the rectangular coordinate system, are the points (a, [#permalink]
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]
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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]
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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]
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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]
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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]
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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: ManhttanGMAT Practice CAT [#permalink]
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
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Re: ManhttanGMAT Practice CAT [#permalink]
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kunalbh19 wrote:

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


There is nothing wrong with your approach. The relation you have got holds (a^2 - b^2 = c^2 - d^2).
But if you start doing algebraic manipulations on the starting point without an eye on what you need to achieve at the end, you may not obtain the result you want. You can manipulate an expression in many ways to get seemingly different results.

Notice that Bunuel got a^2 = c^2 and b^2 = d^2. He could have chosen to add them as
a^2 + d^2 = c^2 + b^2 (or subtract them) which is same as your result. But he chose to add them as a^2 + b^2 = c^2 + d^2 to get the expression he desired. Using the same two expressions, Bunuel arrived at a^2 + b^2 = c^2 + d^2 and you arrived at a^2 - b^2 = c^2 - d^2, both of which are correct. But only one of them (a^2 + b^2 = c^2 + d^2) helps you answer the question directly since you know that the distance of a point (a, b) is given by square root of (a^2 + b^2).

Try to find the relation between individual variables instead of working on the equations as a whole. Either use number plugging or Bunuel's algebraic approach.
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Re: In the rectangular coordinate system, are the points (a, [#permalink]
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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, b) [#permalink]
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1+2:

1) Ensures that the lines joining each of the two points to origin have same slope.
2) Ensures that absicca and ordinates correspondingly have equal magnitudes.
{If one line has points (a,b) then the other line will have coordinates (ak,bk) but (2) ensures that |k| = 1.}

Hence C.

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Re: ManhttanGMAT Practice CAT [#permalink]
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.

Hi Bunnel
i am not understanding how \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\)
have come?
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Re: ManhttanGMAT Practice CAT [#permalink]
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mun23 wrote:
\(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\)
have come?


By convention, only positive roots are considered (at least in GMAT)

\(\sqrt{4} = 2\) (and not -2)

Similarly, \(\sqrt{a^2} = |a|\) i.e. only the positive value
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