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Re: wats this mean? [#permalink]
28 Aug 2008, 08:45

1

This post received KUDOS

1

This post was BOOKMARKED

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 } _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Last edited by x2suresh on 29 Aug 2008, 05:27, edited 1 time in total.

Re: wats this mean? [#permalink]
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?
_________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Re: wats this mean? [#permalink]
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 }

Re: wats this mean? [#permalink]
29 Aug 2008, 00:44

Expert's post

1

This post was BOOKMARKED

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

Re: wats this mean? [#permalink]
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)

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
13 May 2014, 23:51

1

This post received KUDOS

Expert's post

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.