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ps:Geometry

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ps:Geometry [#permalink] New post 06 Apr 2005, 12:40
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
pls explain:

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 [#permalink] New post 06 Apr 2005, 14:50
Is it 2?

The vertices (0,6),(6,2) are the vertices of the diagonal of the square.
Drop a perpendicular to the Y-axis from (6,2) which would be (0,2)(this is one of the vertices of the square)

So the distance from (0,0) to (0,2) is 2.
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Re: ps:Geometry [#permalink] New post 06 Apr 2005, 16:40
Agreed with the answer that the distance from the origin to the closest vertex of the figure is 2.

But the figure is not a square because its sides are not equal. it should be a ractangular to match the answer.
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Re: ps:Geometry [#permalink] New post 06 Apr 2005, 16:49
HIMALAYA wrote:
Agreed with the answer that the distance from the origin to the closest vertex of the figure is 2.

But the figure is not a square because its sides are not equal. it should be a ractangular to match the answer.


It doesnt matter right? One vertex is already on Yaxis which is given and the second we are drawing a Perpn'lar to the Y axis.
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Re: ps:Geometry [#permalink] New post 06 Apr 2005, 16:54
gmat2me2 wrote:
It doesnt matter right? One vertex is already on Yaxis which is given and the second we are drawing a Perpn'lar to the Y axis.


i think so. but the question is flawed and the original originator of the question should know about it.
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 [#permalink] New post 06 Apr 2005, 17:46
Distance from (0,0) to (0,6) = 6
Distance from (0,0) to (6,2) =sqrt(0-6^2 + 0-2^2) = sqrt(36+4)
= 2* sqrt(10) > 6(approximately)
Thus, shortest distance = 6
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 [#permalink] New post 06 Apr 2005, 19:13
draw out the square and the nearest vertex is (0,2) so the distance is 2 form the origin (0,0)
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 [#permalink] New post 07 Apr 2005, 09:19
It it is a square then the closet vertax would not be on Y axis. The side will not be 4, or 6. It'll be 2sqrt(6). We could draw both diagonals and determine the coordinates for the vertax that way. But I'm not going to do the calculations. :P

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 [#permalink] New post 07 Apr 2005, 14:56
HongHu wrote:
It it is a square then the closet vertax would not be on Y axis. The side will not be 4, or 6. It'll be 2sqrt(6). We could draw both diagonals and determine the coordinates for the vertax that way. But I'm not going to do the calculations. :P


The question itself is wrong then. I would take the closest vertex of whatever the intersection of the line from (6,2) and (0,6) would give. They meet on (0,2) .
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 [#permalink] New post 07 Apr 2005, 15:31
gmat2me2 wrote:
HongHu wrote:
It it is a square then the closet vertax would not be on Y axis. The side will not be 4, or 6. It'll be 2sqrt(6). We could draw both diagonals and determine the coordinates for the vertax that way. But I'm not going to do the calculations. :P


The question itself is wrong then. I would take the closest vertex of whatever the intersection of the line from (6,2) and (0,6) would give. They meet on (0,2) .


Alright. I didn't understand your solutions, but they weren't really correct according to my calculations, so here goes.

The two diagonals of a square are at right angles. Their point of intersection bisects the diagonals. Therefore, the point of intersection of the other diagonal is the center of 0,6 and 6,2 = 3,4.
The slope of this diagonal = -2/3.
Therefore the slope of the other diagonal = 3/2.
Using y = 3/2x + c and seeing that 3,4 lies on it, we get c = -1/2.
This gives the equation of the other diagonal as
2y = 3x -1.
We need a point x,y which is 3 away from 3,4 and lies on 2y = 3x -1.
You do the math, but 0,2 doesn't lie on this line, nor is at a distance of 3 from 3,4.

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 [#permalink] New post 07 Apr 2005, 16:44
SQRT(2)

The point of the vertex is (1,1) and the distance will be SQRT(2). Two lines from the vertex to (0,6) and (6,2) will meet at right angle that gives us one equation and also the distance from the vertex to both these points are same (as it is a square) that's another eqn, solving u get (1,1)
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 [#permalink] New post 08 Apr 2005, 01:36
more or less

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 [#permalink] New post 08 Apr 2005, 01:53
Oh!! Boy,

The question is not incorrect. The first co-ordinate (0,6) will correspond to the top left co-ordinate of the square, and the second co-ordinate (6,2) corresponds to the bottom right point of the square. That makes up your diagonal. The other two points will be (6,6) which is the top right hand co-ordinate of the square, and the fourth co-ordinate will be (0,2), which will be the bottom left co-ordinate of the square. Once you have drawn the square, you will see that the co-ordinate that is the closest to the point of origin is (0,2), and therefore the answer is 2.

:twisted:
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 [#permalink] New post 08 Apr 2005, 02:05
Ooops!!

Sorry. I am late to this game. You guys are right. The diagonal doesn't size up.
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 [#permalink] New post 03 Jun 2005, 21:42
I understand how the equation of the other diagonal is 2y = 3x - 1, but I am not sure how you get the co-ordinates of the vertex closest to the origin from here!.

Without knowing the slopes of the edges of the square and without using trignometry, how does one find the specific equations of perpendicular lines and then solve to get the intersection point?.

Any one with the steps to solve this one?...
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Re: ps:Geometry [#permalink] New post 24 Jun 2008, 13:52
@ Prep-GMAT


The important fact is that it is a SQUARE. The diagonals of a square are congruent and perpendicular, so we only have to pay attention to the diagonals.

The easy way is to notice that a line segment from 0,6 to 6,2 has a change of 4 units "DOWN", and six units "RIGHT". So the halfway point would be at 2 units down, and three units right FROM (0,6), giving you (3,4). The halfway point is also the point where the other diagonal intersects the given one, since diagonals of a square bisect each other.

Since the diagonals of a square are equal, we can figure out the other two vertices by counting up/down and left/right from the point where the diagonals intercect, (3,4). This part is a little tricky, but basically, the other two (not given) vertices will both be 3 units away from (3,4) in the Y direction, and two units away from (3,4) in the X direction. This gives (5, 7) and (1, 1) as the other vertices. Figuring out whether to add or subtract from point (3,4) takes some intuition, but the end result is a second diagonal that is equal to, and perpendicular (since the slope of a line segment from 1,1 to 5,7 is +3/2, the negative reciprocal of the slope of line segment 0,6 to 6,2, which is -2/3).

This allows you to find the closest vertex to be 1,1 without any complex math, and it actually can be done quickly.
Re: ps:Geometry   [#permalink] 24 Jun 2008, 13:52
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