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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

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, 06:27, edited 1 time in total.

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

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 \(sqrt {u^2+v^2 }\)

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

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]

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14 May 2014, 00:51

3

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.

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

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04 Dec 2014, 03:27

1

This post was BOOKMARKED

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.

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

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04 Dec 2014, 04:40

Expert's post

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.

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.

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

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19 Feb 2016, 13:59

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