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### Square

by crackgmat007 » Wed Nov 04, 2009 4:46 pm
If 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?

OA - Sqrt 2

Adding the link for explanation that I found very helful. Also, looks like the question was wrong. Updated with the correct info.

https://gmatclub.com/forum/coordinate-ge ... 54747.html
Last edited by crackgmat007 on Thu Nov 05, 2009 10:27 am, edited 3 times in total.

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### ans

by crackthetest » Wed Nov 04, 2009 8:33 pm
It is the distance between (0,0) and (0,2) (use the dist. formula you get answer as 2)

How did you get OA as sqrt(2)?

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### Re: ans

by crackgmat007 » Wed Nov 04, 2009 8:59 pm
crackthetest wrote:It is the distance between (0,0) and (0,2) (use the dist. formula you get answer as 2)

How did you get OA as sqrt(2)?
OA is given as sqrt(2). I am not clear about the explanation.

The mistake in your calculations is that you considered one of the vertex of the square. But the question is asking which vertex is closer to origin. It looks like there is a vertex that is closer than (0,2)

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by NikolayZ » Wed Nov 04, 2009 10:41 pm
if (0;2) is the left point of diagonal, and (6;2) the right one. Then the length of one diagonal is 6, right?
The area of the square then would be 1/2*6*6=18.
then, the length of each side of a square is sqrt(18) or 3sqrt(2).
because the one diagonal is horisontal, the other one must be vertical in square. hence we could find the (x;y) of every vertice.
If so, then the x-measurement of the lowest vertice will be 3, y =(-1).
Then x(3;-1), the distance to zero will be sqrt(9+1)=sqrt(10)...
I can't figure out an answer.

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by crackgmat007 » Thu Nov 05, 2009 10:14 am
Here is the explanation I found in another forum
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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