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

1.u+ v2(read as v square)= -1 2.v<0

Only Explanations and approach please .I know the answer.

I got B, what's the OA?

If you look closely at the question, origin of the coordinate plane can only be closer to (u,v) than (u,v+1) when v is positive. So if we find out about the sign of v, we can determine if (u,v+1) is closer or not.

(1) u + v^2 = -1. In this case, we know u is positive, but nothing about v. INSUFFICIENT.

(2) Knowing that v<0 answers the question. In this case, (u,v+1) is actually closer to the origin than (u,v). SUFFICIENT.

We dont need to know a lot to answer this question. first thing is to look at a graph. We kow that it is going to be a straight line, and that a line is the shortest distance between two points.

Now if we look at a graph, and set V to O...and look at a line for any value of x. If 1 is added to Y we see that the line gets longer (becomes the hypotenuse of the triangle formed in the two lines) however if one is taken away from y the opposite is true, so we need to figure out which direction Y is heading, toward the X axis or away from the x axis.

If x heads toward the x axis it will get shorter if it moves away from the x axis it will get longer...

If V is less than -1/2 the line will always get shorter..V>-1/2 the line will get longer.

I Insufficient, it only tells us that X is negative
II turns out to be also insufficient since 0>V>-1/2 is also possible.

together is also insufficient since 0>v>-1. For instance V = -1/2 and U=-5/4

I thought B at first too, but then i realized that v can be between -1/2 and 0. in this case, (u,v) is closer to the origin than (u,v+1) is.

S1+S2 only tells us that u is negative, but we can still have the same situation. if v is very small, ie -10, then (u,v+1) is closer to the origin than (u,v). but just like before, if v is between 0 and -1/2, then (u,v) will be closer than (u,v+1).

Well, if the (1) statement is changed, the answer should be D. Both equations imply that v is negative and hence (u^2 + v^2) is larger than (u^2 + (v+1)^2) in both cases.
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