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

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Re: Coordinate plane [#permalink] New post 26 Oct 2010, 03:36
IanStewart wrote:
Kronax wrote:
Nice. That's probably the most easiest and efficient way to solve this problem. Kudos from me!


You can actually answer the question without calculating any distances at all (except at the very end), provided you have a good understanding of the meaning of perpendicular slopes. The background: perpendicular slopes are negative reciprocals. If one line has a slope of a/b, a perpendicular line has a slope of -b/a. Now, if a line has a slope of a/b, that is just the ratio of the 'vertical change' to the 'horizontal change' of the line as you move to the right; if a line has a slope of a/b, that means if you move b units to the right, the line will rise by a units. On a perpendicular line, to travel the same distance, you'd *reverse* the horizontal and vertical changes (because the slopes are reciprocals) and also reverse the direction of one of the two changes (because of the negative sign). That is, on a perpendicular line, we'd travel the same distance if we moved right by a units and moved down by b units, or if we moved left by a units and up by b units.

So, in this question, we have a square. The diagonals of a square are perpendicular, and meet at their midpoints. The midpoint of the diagonals in this question is (3,4). Moving from (0,6) to (3,4), we go right by 3 units and down by 2 units. On the perpendicular diagonal, starting from the midpoint, to travel the same distance we need to either go right by 2 units and up by 3 units (to get to (5,7)), or left by 2 units and down by 3 units (to get to (1,1)). So the closest vertex to the origin is (1,1) and the distance is sqrt(2).



The diagonal length is 4\sqrt{13}
Mid point of diagonal is 3,4
Slope of diagonal connecting given points = -2/3
Slope of diagonal connecting require points = 3/2

x1= x + r*cosA and x2 = x - r*cosA - > polar form of line.

y1= y + r*sinA and y2 = y - r*sinA

Since TanA = 3/2 cosA = \frac{2}{\sqrt{13}}

and sinA = \frac{3}{\sqrt{13}}

Using the above equations we get x,y = 1,1 and 5,7

Thus the answer is sqrt2
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Re: Coordinate plane [#permalink] New post 26 Oct 2010, 05:49
Quote:
So, in this question, we have a square. The diagonals of a square are perpendicular, and meet at their midpoints. The midpoint of the diagonals in this question is (3,4). Moving from (0,6) to (3,4), we go right by 3 units and down by 2 units. On the perpendicular diagonal, starting from the midpoint, to travel the same distance we need to either go right by 2 units and up by 3 units (to get to (5,7)), or left by 2 units and down by 3 units (to get to (1,1)). So the closest vertex to the origin is (1,1) and the distance is sqrt(2).


My two cents to a nice explanation:

To give a little info between the two choice - going right by 2 and down by 3 or going right by 3 and down by 2:

Going right by 3 and down by 2 will not result in a square, but a rectangle. This is because then this point and the the end point of the other diagonal (6,2) will end up on the same y coordinate (giving us a quadrilateral that is parallel to X and y axis). Some ppl might have noticed in the beginning that this square can not be straight but slightly revolved in the anti-clock direction.
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Coordinate geometry [#permalink] New post 04 Apr 2011, 22:39
Guys
Pardon me if this discussed before. I am posting this for two reasons -

1. The importance of breathing right on the desk - If you are not breathing right then you may get the hardest questions wrong. I can't overstate the importance of this. When I was in the treadmill the solution clicked effortlessly ha ha. May be this due to abundance of oxygen in the body and the alertness of the brain.

2. Any question can be cracked in 120 secs or less on gmat.

If you exceeded 2 mins then -

1. Your approach is not correct. In this case stop the timer and re-start from a fresh angle.

or

2. You have overanalyzed the problem. Overanalysis always underpays.


3. The gmat iPAD toolkit is the awesome way to streamline your speed.

Without much ado! Now try this ....I will post the 60 sec solution soon.


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)
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Re: Coordinate geometry [#permalink] New post 05 Apr 2011, 17:44
Well another question is -

Does gmat care about exact math? Or I can do some approximations without flouting the accuracy much since I have a stop watch to look :-)
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Re: co-ordinate plane [#permalink] New post 05 Aug 2011, 06:54
Cant think of a easier way than this.. answer is sqrt 2

lets say the vertex is (x,y) ..

--- slope of the diagonal = -2/3 {{ formula is (y2-y1) / (x2-x1) }}
--- midpoint of the diagonal = 3, 4 {{ formula is (x1+x2)/ 2, (y1+y2)/2 }}
--- Slope of other diagonal which passes thru the midpoint and vertex which we are trying to located = (4-y)/ (3-x)

these shud be perpendicular ...
so (4-y)/ (3-x) = 3/2 => 2y = 3x-1 or y= (3x-1)/2

as this is a square the distance from all vertices to midpoint shud be equal.. distace between (0,6) and (3,4) = sqrt 13
distance between (x,y) and (3,4) = sqrt ( (x-3)^2 + (y-4)^2) = sqrt 13
substituting y = (3x-1) / 2 and solving we get x-3 = sqrt 4 => x= 5 or 1

vertices shud be (1,1) and and (5,7)
(1,1) is closest to (0,0) so the distance is sqrt 2

Catch: I have considered the distance from vertex to midpoint to be equal to solve for x and y.. there is one more possible equation, sides shud be of equal length in a square.. but the problem here is that all point on 2y = 3x - 1 will be at an equal distance from both given vertices (0,6) and (6,2).. but not neccesarily perpendicular.. which will form a rhombus and not a square..
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Re: Math Challenge [#permalink] New post 25 Jul 2012, 02:54
KillerSquirrel wrote:
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


Once one of the diagonals is given, your square is uniquely determined. The other diagonal bisects the first one and it is perpendicular to it.
And most important, the sides of the square should not necessarily be and are not (in this case) parallel to the axes.

Start and draw it: first connect the two points to get the first diagonal. Then take the middle of the line segment and draw a perpendicular to the given diagonal. Finally, take two points at equal distances from the middle point and such that the distance between them is the same as the length of the initial diagonal.
Now, you have all the vertices of your square.
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Re: m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal [#permalink] New post 15 Feb 2013, 02:47
this is impossible to solve within 3 mins also in my opinion
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Re: [#permalink] New post 20 Dec 2013, 15:19
hobbit wrote:
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)



take the y difference of the given points, and make it the x difference, and vice versa, then go to the right direction:

Could you please elaborate more on this method. I don't quite get what you are doing here

Cheers!
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Re: m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal [#permalink] New post 18 Apr 2014, 02:37
We can first find the the distance between the end points of the diagonal,i.e square root of 52.
The sides sides of a square are equal so the side of a squre is root 26.(45-45-90)
we know that one point is (6,2), so we can find the the other point which will give us the distance as root 26.
Since we need to minmize the distance use the points(1,1)(the only points applicable in this scenario)
Now we have the required points to calculate the answer.
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m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal [#permalink] New post 05 Aug 2014, 07:54
gmat1220 wrote:
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


I feel this is the only approach that will come close to solving within 2.5 mins. My only problem was estimating final answer. I know sqrt(10) is 3.16 and sqrt(16) is 4. So chosing between 1.73 = sqrt(3) and 1.41 = sqrt(2) could be tricky unless you calculate within +/- 0.3!!
m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal   [#permalink] 05 Aug 2014, 07:54
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