Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
givemeanid
 
 

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

by givemeanid Sun Aug 12, 2007 4:46 pm

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)
Anadi
 
 

Answer should be C

by Anadi Sun Aug 12, 2007 5:48 pm

(1) a/b = c/d

(2) sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)

Ok,

1 is clearly not sufficient. Let's write 2 as

|a|+|b| = |c|+|d| , this is not sufficient since it can have many solutions. Now multiply both sides with |d|

|a||d|+|b||d| = |c||d|+|d||d|..................call it 3

Fom 1, ad=bc, so |a||d| = |b||c|

Replacing in our equation 3

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

|b| (|c|+|d|) = |d| (|c|+|d|)

Hence, |b| = |d|, and using 1 with this, |a| = |c|

So a,b and c,d have same values of co-ordinates if signs are ignored.

Since ditance from origin is square root of sum of squares of x and y co-ordinates, these 2 are equidistant from origin.

So, C is the answer.
givemeanid
 
 

by givemeanid Mon Aug 13, 2007 10:03 am

You da man bro. C is right. I got lazy and didn't factorize enough. Multiplying by |d| was brilliant.
Guest
 
 

Another way to solve??

by Guest Tue Aug 21, 2007 11:30 am

Is there another way to solve the problem above, without introducing the absolute values / intense algebra?

Thanks.
anadi
 
 

I can try

by anadi Tue Aug 21, 2007 1:28 pm

a/b=c/d assume = x

so a=bx, c=dx

Replacing in 2, sqrt(b^2 * x^2) + sqrt(b^2) = sqrt(d^2 * x^2) + sqrt(d^2)

or, (x+1)*sqrt(b^2) = (x+1)*sqrt(d^2) which is, in turn, sqrt(b^2) = sqrt(d^2),

again from (2), sqrt(a^2) = sqrt(c^2), that means b^2 = d^2 and a^2 = c^2

Add these 2, and get a^2 + b^2 = c^2 + d^2, hence both are equidistant.
One more
 
 

Analytical solution

by One more Tue Aug 21, 2007 3:10 pm

Cosider points a,b and c,d in real xy plane. Suppose you are at origin.

2) means that you if you walk on x axis and then turn and go parallel to Y-axis and reach the point, the total distance covered will be same whether you walk to a,b or c,d.

1) Tells you that the ratio of distance on X-axis to that parallel to Y , to reach a,b, is same to the ratio of distance on X-axis to that parallel to Y , to reach c,d.

Now we have 2 pieces of ino, 2 same distances, divided in same ratio, that means respective parts after division are same. hence,

|a| = |c|
|b| = |d|
anadi
 
 

I wonder

by anadi Tue Aug 21, 2007 3:14 pm

Why sometimes it does not show the name when I post mesages, also I am a registered member and have bought access to Question bank and challenges, but can never use id/pwd for forums.
StaceyKoprince
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by StaceyKoprince Tue Aug 21, 2007 7:13 pm

The two systems are different - you can use the same ID and password if you want for both systems, but you have to set it separately yourself for the forums vs. the private stuff on the web site. The forums are free to all users, while the private stuff obviously isn't.

Also - I know the login expires after a while and you have to make sure you're logged in for your ID to post, but I'm not sure if that's why your ID sometimes doesn't post.
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Re: In the rectangular coordinate system, are the points (a, b)

by allen.dsouza Wed Dec 21, 2011 1:38 pm

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|

So we can cancel the third term out from LHS and RHS of equation 2. to get the desired equation
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Re: In the rectangular coordinate system, are the points (a, b)

by tim Tue Jan 10, 2012 7:56 pm

please let us know if you have a question here. otherwise, thank you for sharing your thoughts..
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Re: In the rectangular coordinate system, are the points (a, b)

by chitturiratna Thu Apr 11, 2013 4:31 am

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

I think that according to the question is the distance between

(0,0) and (a,b) is equal to distance between (0,0) and (c,d).

Ans. If The distance between two points is sqrt[(0-a)^2+(0-b)^2] then according to me B is the correct answer.

Please explain the wrongs in my reasoning.
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Re: In the rectangular coordinate system, are the points (a, b)

by RonPurewal Fri Apr 12, 2013 5:37 am

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

I think that according to the question is the distance between

(0,0) and (a,b) is equal to distance between (0,0) and (c,d).

Ans. If The distance between two points is sqrt[(0-a)^2+(0-b)^2] then according to me B is the correct answer.


you appear to be assuming that
√(this + that)
is the same as
√(this) + √(that).

that's not true. (if you don't see why not, just try numbers: e.g., √(16 + 9) = 5, but √16 + √9 = 7.)

in the case of this problem, your error lies in thinking that
√(a^2 + b^2)
and
√(a^2) + √(b^2)
are the same. they aren't.


Please explain the wrongs in my reasoning.


fyi, you can use "mistakes" or "errors" here, but not "wrongs".
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Re: In the rectangular coordinate system, are the points (a, b)

by ankitmohla Sat Dec 14, 2013 7:33 am

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

After reading "equidistant from origin", I presumed a circle with center (0,0) and thought that I had to prove whether (a,b) & (c,d) lie on that circle (whose equation with center (0,0) can be written as x^2 + y^2 = 0)

Therefore, for (a,b) & (c,d) to lie on that circle....I must prove a^2 + b^2 = c^2 + d^2 = 0

Statement 1: Clearly insufficient to derive the above result.

Statement 2: I whole squared both sides (LHS)^2 = (RHS)^2

which implies:--

a^2 + b^2 + 2*sqrt(a^2.b^2) = c^2 + d^2 + 2*sqrt(c^2.d^2)

i.e. a^2 + b^2 + 2*|a|.|b| = c^2 + d^2 + 2*|c|.|d|

from here on I got stuck....because from statement 1, I only know that:
a.d = b.c

So chose (e), because I couldn't prove:
a^2 + b^2 = c^2 + d^2

Dear experts...please show the further path & mistake in my thinking / approach.

*Just for Information: Ron & Stacey, I am big fan of yours...and really grateful to you for your articles and all. They have played major role in my preparation for GMAT.

Also, Is there any specific sub forum? where I can ask for tips & strategies for the last week before actual GMAT.
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Re: In the rectangular coordinate system, are the points (a, b)

by RonPurewal Sun Dec 15, 2013 7:52 am

Yeah, but you don't only know a/b = c/d. By the time you get to using both statements together, you know that a/b = c/d AND you know that |a| + |b| = |c| + |d| (i.e., the first statement).

If a & b are in the same proportion as c & d, try finding some actual numbers that satisfy |a| + |b| = |c| + |d|. For instance, temporarily make a/b = c/d = 1/2, and notice what kinds of numbers actually make both things work. You'll quickly see what's going on here.
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Re: In the rectangular coordinate system, are the points (a, b)

by shikhers747 Wed Nov 12, 2014 10:18 am

RonPurewal Wrote:Yeah, but you don't only know a/b = c/d. By the time you get to using both statements together, you know that a/b = c/d AND you know that |a| + |b| = |c| + |d| (i.e., the first statement).

If a & b are in the same proportion as c & d, try finding some actual numbers that satisfy |a| + |b| = |c| + |d|. For instance, temporarily make a/b = c/d = 1/2, and notice what kinds of numbers actually make both things work. You'll quickly see what's going on here.



>>>>>>>>>>>>>>>>.
Hey, ron it does not matter whether there is a sign difference b/w |a|+|b|=|c|+|d|,
distance will always be sqrt(a2+b2)= sqrt(c2+d2)
2) should be d answer....plz explain if I m wrong