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

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05 Apr 2011, 07:45
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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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05 Apr 2011, 10:29
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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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21 Mar 2007, 15:30
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?

What am I missing on this?
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21 Mar 2007, 15:32
kyatin wrote:
Just reinserting this from an earlier post. This never got discussed.
Quote:
http://www.gmatclub.com/phpbb/viewtopic.php?p=306016#306016

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.
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21 Mar 2007, 15:44
X & Y and mavery - its not rectangle its SQUARE.
Your coordinates of vertices are wrong.
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21 Mar 2007, 16:11
Hmmn...well I too suck in Verbal. Its high time I should stop visiting Math forum and focus on Verbal.

Good luck to you.
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21 Mar 2007, 16:31
kyatin wrote:
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

Good Luck too
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22 Mar 2007, 09:50
Quote:
Hence Slope=3/2 = (x-3)/(y-4) --------(1)

Suithink, I'm not doubting you, I'm just trying to figure this out...Since slope is (y2-y1)/(x2-x1), should this be (y-4)/(x-3)=3/2?

Then if we leave a=x-3 and b=y-4, we get a^2+b^2=13...then divide by a^2, we get 1+(b/a)^2=13/(a^2)...by substitution from (1) =>

1+(3/2)^2=13/(a^2)=>a=+/-2...etc...

Or does it really matter as long as you divide the last equation by the correct variable...
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22 Mar 2007, 09:53
How can this be a sqaure? Knowing the endpoints that we do, the height must be 4, and then width must be 6
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22 Mar 2007, 13:30
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.
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28 Mar 2007, 13:20
Quote:
Equalise them will end up in the equation: 3x-2y =1

I feel kind of silly, but I got lost here. If it's not too much trouble, can you expand a little?
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23 Apr 2007, 05:04
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
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30 Oct 2007, 01:34
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
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30 Oct 2007, 10:10
bkk145 wrote:
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?
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Re: c 15.19 Vertex and distance [#permalink]

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16 Nov 2007, 12:15
bmwhype2 wrote:
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

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)

http://www.gmatclub.com/forum/t54747?si ... dcf6b55b77
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19 Dec 2007, 22:42
bkk145 wrote:
beckee529 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

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?
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19 Dec 2007, 22:44
sondenso wrote:
bkk145 wrote:
beckee529 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

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.
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20 Dec 2007, 19:46
bmwhype2 wrote:
why? where are u getting 3 and 2 from?

Okay,

the slope = (y2-y1)/(x2-x1) = (2-6)/(6-0) = -4/6 = -2/3
The perpendicular diagonal will have a slope of 3/2

Finally, we play with the midpoint:
Vert1 = (3-2),(4-3) = (1,1)
Vert2 = (3+2),(4+3) = (5,7)

The problem is an absolute madness, I was totally lost.

bkk145 - thanks a lot, man!
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20 Dec 2007, 22:48
Why and how is the vertex of the diagonal mentioned, (3,2)?

Could someone please explain the aforementioned statement?
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28 Jul 2008, 20:37
But this doesnt seem to be a square... It could have been one if one of the endpoints were (6,0) instead of (6,2)
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Re: Coordinate   [#permalink] 28 Jul 2008, 20:37

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

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