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# On the coordinate plane is point (0, 0) closer to point (u,

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On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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28 Aug 2008, 08:27
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On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 May 2014, 23:50, edited 4 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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28 Aug 2008, 08:45
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arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE A CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = $$sqrt {u^2+v^2 }$$
Distance between (0, 0) and (u, v+1) = $$sqrt {u^2+v^2 +2v+1}$$

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
$$sqrt {u^2+v^2 +2v+1}$$ is alwasy less than $$sqrt {u^2+v^2 }$$
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Last edited by x2suresh on 29 Aug 2008, 05:27, edited 1 time in total.
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28 Aug 2008, 08:55
hmmm . k .. suresh !! thanks...
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28 Aug 2008, 10:33
x2suresh wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = $$sqrt {u^2+v^2 }$$
Distance between (0, 0) and (u, v+1) = $$sqrt {u^2+v^2 +2v+1}$$

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
$$sqrt {u^2+v^2 +2v+1}$$ is alwasy less than $$sqrt {u^2+v^2 }$$

Why will S2 not work? What is the OA?
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28 Aug 2008, 10:50
leonidas wrote:
x2suresh wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = $$sqrt {u^2+v^2 }$$
Distance between (0, 0) and (u, v+1) = $$sqrt {u^2+v^2 +2v+1}$$

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
$$sqrt {u^2+v^2 +2v+1}$$ is alwasy less than $$sqrt {u^2+v^2 }$$

Why will S2 not work? What is the OA?

Hey Nemo,

2v+1 --> -ve or +ve.
v<0 v=-1/4 2v+1>0
v=-2 2v+1<0
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28 Aug 2008, 12:26
x2suresh wrote:
leonidas wrote:

Why will S2 not work? What is the OA?

Hey Nemo,

2v+1 --> -ve or +ve.
v<0 v=-1/4 2v+1>0
v=-2 2v+1<0

Got it, didn't try a fraction
Thanks X2Suresh
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29 Aug 2008, 00:44
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Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|
Eventually, we can write: v>-0.5

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

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29 Aug 2008, 02:44
thanks , but
how did u get 2 from 1??
could not quite understand...

walker wrote:
Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|......(1)
Eventually, we can write: v>-0.5....(2)

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

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29 Aug 2008, 03:13
Full solution:

|v|<|v+1|

1) v<-1: -v<-v-1 --> 0<-1 --> always false
2) -1<=v<=0: -v<v+1 --> v>-0.5 --> -0.5<v<=0
3) v>0: --> v<v+1 --> 0<1 always true.

Therefore, inequality is true when v>-0.5

Fast solution:

large negative v: inequality is false
large positive v: inequality is true
switch point v=-0.5 --> v>-0.5
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29 Aug 2008, 08:11
Duh! I went for D but after reading your solutions I realized I forgot to test fractions.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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13 May 2014, 14:15
An even better way to look at this one:

Is \sqrt{u^2+v^2} > \sqrt{u^2+ (v+1)^2}
\sqrt{u^2+v^2} > \sqrt{u^2+v^2+2v+1}

So \sqrt{u^2+v^2} is equal on both sides so the deciding factor is 2v+1
2v+1>0 or v>-1/2 then the right side of the inequality is the greater one

I sufficiently tells us...u^2= -1 - v...As u^2 can never be negative..we have
1. v > -1
2. v = -1..in both the cases we get the same answer..Suff

II. V can be between 0 and -1/2..or <-1/2 ...Insuff

A it is
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Last edited by JusTLucK04 on 14 May 2014, 00:58, edited 1 time in total.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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13 May 2014, 23:51
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arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So basically the question asks whether the distance between the points $$(0, 0)$$ and $$(u, v)$$ is less than the distance between the points $$(0, 0)$$ and $$(u, v + 1)$$: is $$\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}$$? --> is $$\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}$$? --> is $$u^2+v^2<u^2+v^2+2v+1$$? --> is $$v>-\frac{1}{2}$$?

(1) $$v + u^2 = -1$$ --> $$v=-1-u^2\leq{-1}$$ --> so the answer to the question is NO. Sufficient.

(2) $$v<0$$. Not sufficient.

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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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24 Jun 2014, 21:49
How we could have represented this equation on graph ..... and solved by using graph technique

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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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25 Jun 2014, 01:44
GmatDestroyer2013 wrote:
How we could have represented this equation on graph ..... and solved by using graph technique

This question is not a good candidate for graphic approach.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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04 Dec 2014, 02:27
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Bunuel wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So basically the question asks whether the distance between the points $$(0, 0)$$ and $$(u, v)$$ is less than the distance between the points $$(0, 0)$$ and $$(u, v + 1)$$: is $$\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}$$? --> is $$\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}$$? --> is $$u^2+v^2<u^2+v^2+2v+1$$? --> is $$v>-\frac{1}{2}$$?

(1) $$v + u^2 = -1$$ --> $$v=-1-u^2\leq{-1}$$ --> so the answer to the question is NO. Sufficient.

(2) $$v<0$$. Not sufficient.

hi Bunuel,

i could not understand the last step.

v=-1-u^2\leq{-1} --> so the answer to the question is NO.

I did not get how did you infer 'v' lesser than -1 from the last equation??
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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04 Dec 2014, 03:40
arshu27 wrote:
Bunuel wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So basically the question asks whether the distance between the points $$(0, 0)$$ and $$(u, v)$$ is less than the distance between the points $$(0, 0)$$ and $$(u, v + 1)$$: is $$\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}$$? --> is $$\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}$$? --> is $$u^2+v^2<u^2+v^2+2v+1$$? --> is $$v>-\frac{1}{2}$$?

(1) $$v + u^2 = -1$$ --> $$v=-1-u^2\leq{-1}$$ --> so the answer to the question is NO. Sufficient.

(2) $$v<0$$. Not sufficient.

hi Bunuel,

i could not understand the last step.

v=-1-u^2\leq{-1} --> so the answer to the question is NO.

I did not get how did you infer 'v' lesser than -1 from the last equation??

As for your question, we need to find whether $$v>-\frac{1}{2}$$.

(1) gives $$v=-1-u^2$$. Since u^2 (the square of a number) must be non-negative, then we have that $$v=-1-(nonnegative)\leq{-1}$$, therefore v is NOT greater than -1/2, so we have a definite NO answer to the question.

Hope it's clear.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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