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

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Re: Coordinate geometry [#permalink] New post 05 Apr 2011, 05:03
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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
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Re: Coordinate geometry [#permalink] New post 05 Apr 2011, 06:45
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gmat1220 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?

(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)


Sol:

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]

-\frac{2}{3} * Slope_{(AC)} = -1
Slope_{(AC)} = \frac{3}{2}

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

(x-3)^2+(\frac{3}{2}x-\frac{1}{2}-4)^2=(\frac{\sqrt{52}}{2})^2
(x-3)^2(1+\frac{9}{4})=13
(x-3)^2=4
x=5 \hspace{2} & \hspace{2}x=1

When x=1, y=\frac{3}{2}x-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}=1
When x=5, y=\frac{3}{2}*5-\frac{1}{2}=\frac{15}{2}-\frac{1}{2}=7

Thus, we found the other two vertices of the square viz. (1,1) \hspace{2} & \hspace{2} (5,7)

Vertex (1,1) is closest to (0,0)

Distance between (0,0) \hspace{2} & \hspace{2} (1,1)
Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
Distance = \sqrt{(1-0)^2+(1-0)^2} = \sqrt{2}

Ans: "B"
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Re: Coordinate geometry [#permalink] New post 05 Apr 2011, 09:29
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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


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}
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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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