Problem Solving

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


Please explain.
----------------

Am I missing something, or do those coordinates result in a non-square quad.?

vscid 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


Please explain.
----------------

please double check coordinates. they create non-square.

It will be a square where the sides are NOT parallel to either of the axes.

Cheers.

shahdevine wrote:

vscid 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


Please explain.
----------------

please double check coordinates. they create non-square.

OA is C...

Cant we make a square with sides (0,6) , (1,1) , (6,2) and (5,7)...???

Would you mind explaining how you got to those coordinates?

tom4lax wrote:Am I missing something, or do those coordinates result in a non-square quad.?

Why do you think the coordinates result in a non-square?

endpoints of diagonal (0,6) and (6,2) = difference in x values of 6 and y values of 4, sides aren't of same length. However, if the sides aren't parallel to the axis, then possibly it is a square. I just don't know how you would find these points?

Don't know a quick way to find.. It took me almost 10 min to reach to the solution...


The two points are (6,2) and (0,6)

The length of diagonal = sqrt(52)

Length of the side = sqrt(26)

Now the two points we know and the length we know so using the formula sqrt( sqr(x1-x2) + sqr(y1-y2)) we can get two equations..

Let the 3rd point in square be (x,y)

For (6,2) == x^2 + y^2 -12x - 4y + 14 = 0 ------- 1

for (6,0) == x^2 + y^2 -12y + 10 = 0 ------------- 2

subtract one from other then we get 3x - 2y = 1

x = 1/3(1+2y)

Sub x in 2

we get y^2 - 8y +7 = 0

so values of Y = 1 , 7
then value of X = 1, 5

(1,1) is nearer to (0,0)

so the distance = sqrt(2)

tom4lax wrote:endpoints of diagonal (0,6) and (6,2) = difference in x values of 6 and y values of 4, sides aren't of same length. However, if the sides aren't parallel to the axis, then possibly it is a square. I just don't know how you would find these points?

Nowhere it is mentioned that the sides are parallel to the axis.

Right, thanks, I realize that now.

How did you solve the question?

tom4lax wrote:Right, thanks, I realize that now.

How did you solve the question?

Almost the same way, aspireGMAT did.