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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?

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)â€¦ 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. _________________

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)

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

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.

Re: Math Challenge [#permalink]
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 ?

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

Re: coordinate geometry [#permalink]
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?

Re: c 15.19 Vertex and distance [#permalink]
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)

Re: coordinate geometry [#permalink]
19 Dec 2007, 21:42

bkk145 wrote:

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 [b]3 in the y coordinate and distance of 2 in the x coordinate.[/b]

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)

I have the same question that how you can get information in the boldface? Do you mind showing me more details?

Re: coordinate geometry [#permalink]
19 Dec 2007, 21:44

sondenso wrote:

bkk145 wrote:

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 [b]3 in the y coordinate and distance of 2 in the x coordinate.[/b]

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)

I have the same question that how you can get information in the boldface? Do you mind showing me more details?

Its best to draw it on a graph. Ul see that the midpoint is exactly 3 points away from the x on 6,2 and 2 points away from the y cooridinate.

again try drawing a graph i dunno how else to explain it.

I really doubt it is a gmat level prob. The only way i can think of solving it ...

1). Slope of given diagonal (call it d1) is -2/3. Hence slope of other diagonal (call it d2) is 3/2. 2). Use 3/2 to find the equation of d2. (it comes out to be 3x-2y-1=0). 2). The midpoint of the d1 is (3,4). This should also be the midpoint of d2. 3). Length of the d1 = length of d2 = 2 * (13^0.5) call it D.

Therefore you need two points at distance D from (3,4) on the line d2. This involves use of sin and cos functions on argument z, where tan(z)=slope of d2=3/2. Had the angle z been 30 or 60 or 0 degrees, it would have been fairly simple, but with tan(z)=3/2.

The points would be ( 3+D cos(z), 4+D sin(z) ) and ( 3-D cos(z), 4-D sin(z) ). Find the distance of these from the origin to get your answer.

Hope I am not missing something very simple...

Last edited by bhushangiri on 29 Jul 2008, 04:03, edited 1 time in total.