Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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?

Midpoint of two points \((x_1, y_1) \hspace{2} & \hspace{2} (x_2, y_2)\) \(Midpoint(M_{x1},M_{y1})=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\)

Similarly, let's find the midpoint for \((0,6) \hspace{2} & \hspace{2} (6,2)\) \(Midpoint(M_{x1},M_{y1})=(\frac{0+6}{2}, \frac{6+2}{2})= (3,4)\)

Distance between two points \((x_1, y_1) \hspace{2} & \hspace{2} (x_2, y_2)\) \(Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Similarly, the length of the diagonal is the distance between two points \((0,6) \hspace{2} & \hspace{2} (6,2)\) \(Diagonal = \sqrt{(6-0)^2+(2-6)^2} = \sqrt{36+16} = \sqrt{52}\)

Equation of a line passing through points \((x_1, y_1) \hspace{2} & \hspace{2} (x_2, y_2)\) \(y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

Similarly, Equation of the line segment or diagonal BD passing through \((0,6) \hspace{2} & \hspace{2} (6,2)\) \(y-6=\frac{2-6}{6-0}(x-0)\) \(y-6=-\frac{2}{3}x\) \(y=-\frac{2}{3}x+6\)

The slope of the diagonal \(BD=-\frac{2}{3}\)

Slope of the diagonal BD * Slope of the diagonal AC = -1 [Note: Both diagonals are perpendicular]

Equation of a line with slope \(\frac{3}{2}\) passing through point \((3,4)\) \(y-y_1 = m(x-x_1)\) \(y-4 = \frac{3}{2}(x-3)\) \(y = \frac{3}{2}x-\frac{1}{2}\) -----------------1

Now, if we take \((3,4)\) as the midpoint of a circle with radius=half of diagonal of the square; Then, \(radius = \frac{\sqrt{52}}{2}\)

Equation of the circle will be: \((x-x_1)^2+(y-y_1)^2=(radius)^2\) \((x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2\) -------------2

We can solve equation 1 and 2 to get the value for x and y because the circle will meet the diagonal AC at two points

Here you go - The property of square is - diagonals intersect at midpoint and are equal in length. [(6, 2) and (0, 6)] is one diagonal. The mid point of this diagonal is M (6/2,(6+2)/2) i.e M (3,4)

The second diagonal passes through the nearest vertex (lets say A) and point M. Extend the second diagonal AM to meet the origin O(0,0). We have now OM.

OM = OA + AM. We need OA - distance of nearest vertex from the origin AM = Half of diagonal = sqrt(6^2 + 4^2)/2 = sqrt(52)/2 = sqrt(13) OM = sqrt(4^2 + 3^2) = 5 OA = OM - AM = 5 - sqrt(13) = 5 - 3.6 = 1.4 (approx). Hence C . sqrt(2) = 1.4

The solution above assumes that origin lies on the line AM which need not be the case (indeed it is not the case as we determine once we solve for the A coordinate).

Alternatively,

we know that mid-point for the diagonal is \((3,4)\). Now, let the other vertice (lets say A) have the coordinates \((h,k).\)

We know that slope of given diagonal is \((-2/3)\), so slope of other diagonal is 3/2.

We also know that length of diagonal is \(\sqrt{52}\), so length from any one vertice is \(\sqrt{13}\)

now, we know \(\frac{(4-k)}{(3-h)} = \frac{3}{2}\) (by slope formula)

and \((4-k)^2+(3-h)^2=13\) (by distance formula).

substituting \((4-k) = 3/2*(3-h)\) in above, we get

\(13/4*(3-h)^2 = 13\) or \(h=1\) or \(h=5\) and hence \(k=1\) or \(k=7\)

So, the vertice closer to origin is \((1,1)\) and distance is \(\sqrt{2}\)

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. _________________

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.

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

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?

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)

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?

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.