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

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In the rectangular coordinate system, are the points (a, b) [#permalink] New post 17 Aug 2011, 10:42
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A
B
C
D
E

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65% (02:46) correct 35% (01:06) wrong based on 49 sessions
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)


OA follows..





OA is C.

I have seen a couple of nice approaches on number plugging, but just checking if there are any other innovative ideas to think conceptually on this. I myself tried algebra...then tried to switch mid-way to plugging and messed it on time...

So pointers on CONCEPT, GUESSING, TAKEAWAYS on how to recognize what approach to adopt,,... all welcome.


Thank you

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Last edited by Bunuel on 07 Apr 2012, 07:28, edited 1 time in total.
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Re: Guaranteed Toughie 700+ - looking for creative ideas ways.. [#permalink] New post 18 Aug 2011, 06:23
mokap25 wrote:
Toughie from MGMAT - looking for creative and different approaches to approach this problem.

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)




Not too difficult I think.

(1) A/B = C/D tells us that a/b & c/d are proportionally the same. Take an example of a/b = 1/2 and c/d = -4/-8 to understand it is not sufficient.

Now (2) is basically saying the absolute value of the sum is the same for both points. Of course this could be true with (2,2) and (4,0) which are not the same distance from the origin.

(1)+(2) Now if you just think about it, two points that have the same ratio, and the same absolute sum must be the same distance from the origin.
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Re: Guaranteed Toughie 700+ - looking for creative ideas ways.. [#permalink] New post 18 Aug 2011, 16:33
mokap25 wrote:
Toughie from MGMAT - looking for creative and different approaches to approach this problem.

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)

OA follows..

OA is C.

I have seen a couple of nice approaches on number plugging, but just checking if there are any other innovative ideas to think conceptually on this. I myself tried algebra...then tried to switch mid-way to plugging and messed it on time...

So pointers on CONCEPT, GUESSING, TAKEAWAYS on how to recognize what approach to adopt,,... all welcome.

Thank you


The question is asking about the distance from the origin.

Statement 1 says only about the ratio (a/b = c/d) of the coordinates of the points (a, b) and (c, d).
If a = 3 and b = 1, a/b = 3.
c/d could be three in many ways...That's clearly not sufficient..

Statement 2:
sqrt(a^2)+sqrt(b^2) =sqrt(c^2) +sqrt(d^2)
i.e. a+b = c+d.

This is possible in many ways. This is also not sufficient..

Statements 1 and 2: I wonder how to solve algebraically!

a/b = c/d ........................(i)
a + b = c + d ....................(ii)

suppose a/b = c/d = 3
a = 3b
c = 3d

3b + b = 3d + d ....................(iii)
4b = 4d
b = d

If b = d, a has to be equal to c. If so, coordinates points (a, b) and (c, d) are in equi-distance from the origin...

C.
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Re: Toughie from MGMAT - looking for creative and different [#permalink] New post 07 Apr 2012, 07:19
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IMO B......

bunnel request your explanation why b alone can't be sufficient......
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Re: Toughie from MGMAT - looking for creative and different [#permalink] New post 07 Apr 2012, 07:26
Expert's post
vdbhamare wrote:
IMO B......

bunnel request your explanation why b alone can't be sufficient......


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: Toughie from MGMAT - looking for creative and different   [#permalink] 07 Apr 2012, 07:26
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