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m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal

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m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal [#permalink] New post 21 Mar 2007, 14:05
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On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) \frac{1}{\sqrt{2}}

(B) 1

(C) \sqrt{2}

(D) \sqrt{3}

(E) 2\sqrt{3}

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 [#permalink] New post 21 Mar 2007, 14:30
I get 2 on this, but I don't see it in the answer choices. If the diagonal end points are (6,2) and (0,6), this puts the closest vertex at (0,2). Therefore the distance is 2?

What am I missing on this?
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Re: PS Geometry. [#permalink] New post 21 Mar 2007, 14:32
kyatin wrote:
Just reinserting this from an earlier post. This never got discussed.
Quote:
http://www.gmatclub.com/phpbb/viewtopic.php?p=306016#306016


On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

a) 1/sqrt (2) b) 1 c) sqrt (2) d) sqrt (3) e) 2*sqrt (3)


If the endpoints of the diagonal of the square are (6,2) and (0,6)… Then the vertices of the square are (0,2)-(0,6)-(6,6)-(6,2).. So the closest vertex to (0,0) is (0,2) and the distance should be 2.
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 [#permalink] New post 21 Mar 2007, 14:35
(C) for me :)

I use a method that is out of scope of the GMAT... I prefer to use the properties of perpendicular vectors... :)

First of all, I search the mid point I that is the center of gravity of the square.

I((6+0)/2 , (2+6)/2) <=> I(3,4)

Lets call:
o A : the vertice at (6,2)
o C : the vertice at (0,6)
o B : the vertice "above" the line AC
o D : the vertice "below" the line AC

Then, I calculate the vectors that we can have:
o Vector(IA) = (6-3,2-4) = (3,-2)
o Vector(IC) = (0-3,6-4) = (-3,2)

Then, as Vector(IA) is perpendicular to Vector(IB) and Vector(ID) is perpendicular to Vector(IC), we have the rules of perpendicular vectors:
o Vector(IB) = (-y(IA) , x(IA)) = (2,3)
o Vector(ID) = (-y(IC) , x(IC)) = (-2,-3)

Then, we rebuild the coordonate of B and D:
o B(3+2 , 4+3) <=> B(5 , 7)
o D(3-2 , 4-3) <=> D(1 , 1)

So, we have
o A(6,2)
o B(5,7)
o C(0,6)
o D(1,1)

Thus, OD is the shortest distance and OD = sqrt(2)
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 [#permalink] New post 21 Mar 2007, 14:44
X & Y and mavery - its not rectangle its SQUARE.
Your coordinates of vertices are wrong.
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 [#permalink] New post 21 Mar 2007, 15:11
Hmmn...well I too suck in Verbal. Its high time I should stop visiting Math forum and focus on Verbal.

Good luck to you.
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 [#permalink] New post 21 Mar 2007, 15:31
kyatin wrote:
Hmmn...well I too suck in Verbal. Its high time I should stop visiting Math forum and focus on Verbal.

Good luck to you.


Ok :)... A balanced practice and a focus on weaknesses is the key :)...

By the way, I'm done with the GMAT for a while now :).... But I still need your wish of good luck ;).... Tomorrow, I will know wheither I'm in HEC :)... So, more than ever... Crossed fingers :D :)

Good Luck too :)
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 [#permalink] New post 22 Mar 2007, 04:11
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Here goes my solution:

Let (x,y) be the unknown vertex...

Slope of given diagonal =(6-2)/(0-6)=-2/3
Hence Slope of other diagonal =3/2
Midpoint of 'other dia' = (3,4)

Hence Slope=3/2 = (y-4) /(x-3)--------(1)--> Mavery see below corrected

distance of vertex (x,y) from (3,4) = sqrt 52/2
i.e (x-3)^2+ (y-4)^2= 52/4=13-----(2)

Let (x-3) =a,(y-4) =b
so eqn (2) becomes a^2+ b^2= 13
Dividing by b^2 throughout
(a/b)^2+ 1 =13/(a^2)
From eqn 1 a/b =3/2
Gives b=+/- 3
i.e y-4 = 3 =====> y= 7
y-4=-3======> y= 1
Similarly a = +/-2
====> x is 5,1
Vertices are (5,7) and (1,1)
Now we are having the two vertices ...clearly the closest one to origin is (1,1) ...
So distance is sqrt(2)...

What is the source of this Q ...?

Last edited by suithink on 22 Mar 2007, 09:43, edited 3 times in total.
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 [#permalink] New post 22 Mar 2007, 08:50
Quote:
Hence Slope=3/2 = (x-3)/(y-4) --------(1)


Suithink, I'm not doubting you, I'm just trying to figure this out...Since slope is (y2-y1)/(x2-x1), should this be (y-4)/(x-3)=3/2?

Then if we leave a=x-3 and b=y-4, we get a^2+b^2=13...then divide by a^2, we get 1+(b/a)^2=13/(a^2)...by substitution from (1) =>

1+(3/2)^2=13/(a^2)=>a=+/-2...etc...

Or does it really matter as long as you divide the last equation by the correct variable...
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 [#permalink] New post 22 Mar 2007, 08:53
How can this be a sqaure? Knowing the endpoints that we do, the height must be 4, and then width must be 6
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 [#permalink] New post 22 Mar 2007, 12:30
Thanks Suithink, I was just making sure I was understanding correctly.

Tuneman, I think you may be thinking like I was at first. I was making the assumption that the "base" of the square was parallel to the x-axis. Try to think of the square as somewhat tilted. So the vertex of the square at the bottom right actually has a greater y-value than the vertex in the lower left corner.
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 [#permalink] New post 28 Mar 2007, 03:49
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My solution:

Let the nearest vertex is (x,y)
the length of the diagonal is = distance between (0,6) and (6,2) = sqrt(52). Hence the length of each side of the square = sqrt(26)
Now the distance between the each of the known vertices and the nearest vertex should be the length of a side. Equalise them will end up in the equation: 3x-2y =1
distance between (0,6) and (x,y) is sqrt(26), solving this y = 1 or 7. Putting the values of y in 3x-2y =1 , possible values of x are 1, 5.
Hence the nearest vertex is (1,1).
distance between (0,0) and (1,1) is sqrt(2).
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 [#permalink] New post 28 Mar 2007, 12:20
Quote:
Equalise them will end up in the equation: 3x-2y =1


I feel kind of silly, but I got lost here. If it's not too much trouble, can you expand a little?
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 [#permalink] New post 23 Apr 2007, 02:34
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answer is C (sqrt(2))

it is clear that the closest vertex is the bottom left one (the bottom right is (6,2) and the upper left is (0,6))
so let's find its coordinates:

first, let's find the middle point (which is half way the diagonal)
since the whole diagonal spans between 0 and 6 on the x-axis the middle point is on 3. and on the y-axis it spans from 2 to 6 so the middle-point is at 4. so the middle point is at (3,4)

the bottom left is 90 degress down and left from the middle point, in the same distance of the end points of the diagonal....
now - here is a nice method to go 90 degrees from a given 2 points:
take the y difference of the given points, and make it the x difference, and vice versa, then go to the right direction:

so the x difference between (6,2) and (3,4) is 3 and y difference is 2
we our target point is with x-difference 2 (from (3,4)) and since it must be on the left - the x axis is 1.
the y difference is 3, and since the point should be on the bottom we substract it and get 1.

so the bottom left vertex is (1,1), and the distance from (0,0) is sqrt(2)
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Re: Math Challenge [#permalink] New post 23 Apr 2007, 04:04
candice.chan wrote:
On the coordinate plance (6,2) and (0,6) are the endpoints of the diagonal of a square. What is the distance between point (0,0) and the closest vertext of the square?

a) 1/sqt(2)
b) 2
c) sqrt(2)
d) sqrt(3)
e) 2*sqrt(3)

Thx!


sorry to ask such a silly question , but can someone please explain how can it be a squere, if the coordinate on the endpoints of the diagonal of a square are (6,2) (0,6) ? dosen't it gives a rectangle area and not a square ?

thanks
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 [#permalink] New post 23 Apr 2007, 22:15
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Answer C - sqrt2

Folowed a slightly different method though,

- by observation, the bottom left point is the closest to the origin, lets name the point (x,y)

now, since the figure is a square, all sides are equal in length.

distance between (0,6) and (x,y) = dist btw (6,2) and (x,y)

this gives the eqn : 3x-2y = 1

Now, by looking closely at the eqn and taking into consideration that the figure is a square, it can be seen that (1,1) solves for x,y

hence distance = sqrt2

Alternative :

to mathematically find the x,y coordinate,

- the product of slopes of perpendicular lines is -1

using the above, product of slope of line segment with end points (6,2) and (x,y) and the line segment with end points (0,6) and (x,y) = -1

this gives us the 2nd eqn.

so we can solve for x,y and arrive at the solution and forget about the remaining 36 questions in the Q section :):)
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 [#permalink] New post 30 Oct 2007, 00:34
To me, we should:
> Draw a quick XY Plane to have a better idea of what is going on
> Know the middle point of A(6, 2) and B(0, 6)
> Determine the nearest vertex to 0(0,0) by looking in which cadran this middle point is
> Create the equation of circle on which all vertice lie on, centered so at the middle point of AB
> Create the equation of the line AB
> Create the equation of the line perpendicular to AB and passing by the middle point
> Calculate the coordonate of the nearest vertex to 0 by using the equation of the circle and the perpendicular line to AB

Another way can be to use an approach with vectors :)
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Re: coordinate geometry [#permalink] New post 30 Oct 2007, 07:10
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beckee529 wrote:
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

1√2
1
√2
√3
2√3

i know the answer and have the explanation. Please explain your approach or logic, thanks.


The mid point between (6,2) and (0,6) is (3,4)
You know that the vertexes of the diagonal mentioned has distance of 3 in the x coordinate and distance of 2 in the y coordinate.
Since both diagonals must be perpendicular, using the perpendicular property (inverse of slope), it must be that those distances are reversed.

This means that from point (3,4), the new vertexes will have distance of 3 in the y coordinate and distance of 2 in the x coordinate.

So two other vertexes will be at:
(3-2, 4-3) = (1,1)
and
(3+2, 4+3) = (5,7)

The closer to the (0,0) is (1,1)
Distance between (0,0) and (1,1) = sqrt(2)
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Re: coordinate geometry [#permalink] New post 30 Oct 2007, 09:10
bkk145 wrote:
The mid point between (6,2) and (0,6) is (3,4)
You know that the vertexes of the diagonal mentioned has distance of 3 in the x coordinate and distance of 2 in the y coordinate.
Since both diagonals must be perpendicular, using the perpendicular property (inverse of slope), it must be that those distances are reversed.

This means that from point (3,4), the new vertexes will have distance of 3 in the y coordinate and distance of 2 in the x coordinate.

So two other vertexes will be at:
(3-2, 4-3) = (1,1)
and
(3+2, 4+3) = (5,7)

The closer to the (0,0) is (1,1)
Distance between (0,0) and (1,1) = sqrt(2)


Man I wish I had the clarity of thinking that bkk145 has. I needed 20 mins to come to that solution.

BKK145 I would like to borrow your brain for the math section on November 21.. only 75 minutes ok?
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Re: c 15.19 Vertex and distance [#permalink] New post 16 Nov 2007, 11:15
bmwhype2 wrote:
On a coordinate plane (6,2) and (0,6) are the endpoints of the diagonal of a square. What is the distance between point (0,0) and the closest vertex?

1/rad2
1
rad2
rad3
2rad3

Please explain your answer.


The mid point between (6,2) and (0,6) is (3,4)
You know that the vertexes of the diagonal mentioned has distance of 3 in the x coordinate and distance of 2 in the y coordinate.
Since both diagonals must be perpendicular, using the perpendicular property (inverse of slope), it must be that those distances are reversed.

This means that from point (3,4), the new vertexes will have distance of 3 in the y coordinate and distance of 2 in the x coordinate.

So two other vertexes will be at:
(3-2, 4-3) = (1,1)
and
(3+2, 4+3) = (5,7)

The closer to the (0,0) is (1,1)
Distance between (0,0) and (1,1) = sqrt(2)




http://www.gmatclub.com/forum/t54747?si ... dcf6b55b77
Re: c 15.19 Vertex and distance   [#permalink] 16 Nov 2007, 11:15
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