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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Coordinate Problem!
This topic has expert replies
-
Frankenstein
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
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.
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.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
-
Ozlemg
- Master | Next Rank: 500 Posts
- Posts: 407
- Joined: Tue Jan 25, 2011 9:19 am
- Thanked: 25 times
- Followed by:7 members
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?
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?
The more you suffer before the test, the less you will do so in the test! 
-
amit2k9
- Master | Next Rank: 500 Posts
- Posts: 461
- Joined: Tue May 10, 2011 9:09 am
- Location: pune
- Thanked: 36 times
- Followed by:3 members
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.
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.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
-
GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Distance from the origin: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?
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
sanabk
- Master | Next Rank: 500 Posts
- Posts: 174
- Joined: Fri Apr 02, 2010 3:41 pm
- Thanked: 6 times
- Followed by:1 members
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!!!
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!!!
-
haqiqatprast
- Newbie | Next Rank: 10 Posts
- Posts: 2
- Joined: Sat Jul 09, 2011 12:01 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.
|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.
