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# M15-19

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:55
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Difficulty:

85% (hard)

Question Stats:

66% (02:16) correct 34% (02:23) wrong based on 85 sessions

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If 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. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:55
5
3
Official Solution:

If 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. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

Given endpoints of diagonal of a square: $$B(0,6)$$ and $$D(6,2)$$. Let other vertices be $$A$$ (closest to the origin) and $$C$$ (farthest to the origin):

Length of the diagonal would be: $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}$$

Coordinates of the midpoint $$M$$ of the diagonal would be: $$M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)$$.

{The slope of line segment AM}*{The slope of line segment BD}=-1 (as they are perpendicular to each other), so $$\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1$$, which simplifies to $$y-4=\frac{3}{2}(x-3)$$

The distance between the unknown vertices to the midpoint is half the diagonal:

$$(x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13$$;

$$(x-3)^2+\frac{9}{4}(x-3)^2=13$$;

$$(x-3)^2=4$$;

Which gives: $$x=1$$ and $$y=1$$ OR $$x=5$$ and $$y=7$$.

Hence point $$A$$ is at $$(1,1)$$ and point $$C$$ is at $$(5,7)$$. Closest to the origin is $$A$$. Distance $$OA=\sqrt{2}$$

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29 Aug 2015, 01:50
1
Hi

Is there any shorter way to solve this problem.
I have understood the solution but this seems to be a very tough and complex problem.
I am wondering if this can be solved in some other way.
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13 Nov 2015, 11:54
1
2
Hi akhil911, maybe this will help. I calculated the midpoint in the same way as described by Bunuel to find (3,4). Then I determined the delta from point B to the midpoint M as (0,6) - (3,4) = (3,2). In order to find point A simply swap the x and y and subtract from the midpoint: (3,4) - (2,3) = (1,1). Which gives √2 as distance.
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26 Dec 2015, 07:19
1
1
Or for points (6,2) and (0,6) in this case:
Absolute difference for x is 6-0=6
Absolute difference for y is 6-2=4
Using logic, since this is a square, sides need to be equal and average of 6+4 is 5
Hence, absolute difference of 5 units away is for x is x=1 and for y is y=1, with intersection at point (1, 1) being equidistant to B and D
Then work out the distance as before
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30 May 2016, 05:12
Shorter solution if one can draw a correct sketch of the figure:

The shortest distance from (0,0) to one of the vertexes of the square is perpendicural distance to A. A and (0,0) forms a square with s=1. The shortest distance will be the diagonal of that square which is V2s or 1 x V2 = V2 the answer.

Even if the graph is drawn somewhat right, we are looking for the answer that has nV2 in it.
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06 Jul 2016, 14:16
The distance between the unknown vertices to the midpoint is half the diagonal:

(x−3)2+(y−4)2=(52√2)2=13(x−3)2+(y−4)2=(522)2=13;

I don't understand what this step is doing?
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27 Jul 2016, 10:59
rachitshah
It seems like the steps from the given explanation accomplish the following:
1) Find the distance of the original diagonal: (6-0)^2 + (2-6)^2 = X^2; X = root 52
2) Find the midpoint of the original diagonal: (6+0)/2, (6+2)/2 = (3,4)
3) Find the equation for the line for the other diagonal, since this will be the line that has the coordinates closest to (0,0)
Key to this step: This line is the perpendicular bisector of the diagonal (the diagonals of squares are perpendicular bisectors = fact in geometry)
You can find the equation for the line the other diagonal by doing the following:
a) y = mx + b (generic equation for line)
b) find the slope of the original diagonal: (6-2)/(0-6) = -2/3 and take the negative reciprocal of this = 3/2 since the other diagonal is perpendicular
c) now you have y = 3x/2 + b; find the value of b by plugging in (3,4) and you get y = 3x/2 - 1/2, which is the same thing as y-4 = 3/2(x-3)
4) Distance from farthest coordinates to midpoint of diagonal = Distance from shortest coordinates to midpoint of diagonal = 1/2 the length of the diagonal (in this case, (root52/2)^2).
5) The distance from the farthest or shortest coordinates of the diagonal to the midpoint can be solved by using pythagorean theorem:
(x-3)^2 +(y-4)^2 = (root52/2)^2. (root52/2)^2 = 13
6) From step 3 you know that y-4 = 3/2(x-3). Plug this into y for step 5 above ==> (x-3)^2 + (y-4)^2 = 13 --> (x-3)^2 + [3/2(x-3)]^2 = 13 -->
(x-3)^2 +9/4(x-3)^2 = 13; solving for x gives you (x-3)^2 = 4, so x can be 1 or 5.
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28 Aug 2016, 09:47
I think this is a high-quality question and I agree with explanation.
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24 Oct 2016, 00:39
rachitshah wrote:
The distance between the unknown vertices to the midpoint is half the diagonal:

(x−3)2+(y−4)2=(52√2)2=13(x−3)2+(y−4)2=(522)2=13;

I don't understand what this step is doing?

This step is finding the distance from vertex A(x,y) to mid point M(3,4) or the length of AM you can say.
since AC is also a diagonal, AM will be half its length (which is\sqrt{52} and diagonals of a square are equal in length.)

I hope it clears your doubt.
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10 Aug 2017, 23:29
Skytail wrote:
Hi akhil911, maybe this will help. I calculated the midpoint in the same way as described by Bunuel to find (3,4). Then I determined the delta from point B to the midpoint M as (0,6) - (3,4) = (3,2). In order to find point A simply swap the x and y and subtract from the midpoint: (3,4) - (2,3) = (1,1). Which gives √2 as distance.

Can you pls explain the logic behind this approach. What do you mean by delta here.
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10 Aug 2017, 23:41
Bunuel wrote:
Official Solution:

If 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. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

Given endpoints of diagonal of a square: $$B(0,6)$$ and $$D(6,2)$$. Let other vertices be $$A$$ (closest to the origin) and $$C$$ (farthest to the origin):

Length of the diagonal would be: $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}$$

Coordinates of the midpoint $$M$$ of the diagonal would be: $$M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)$$.

{The slope of line segment AM}*{The slope of line segment BD}=-1 (as they are perpendicular to each other), so $$\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1$$, which simplifies to $$y-4=\frac{3}{2}(x-3)$$

The distance between the unknown vertices to the midpoint is half the diagonal:

$$(x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13$$;

$$(x-3)^2+\frac{9}{4}(x-3)^2=13$$;

$$(x-3)^2=4$$;

Which gives: $$x=1$$ and $$y=1$$ OR $$x=5$$ and $$y=7$$.

Hence point $$A$$ is at $$(1,1)$$ and point $$C$$ is at $$(5,7)$$. Closest to the origin is $$A$$. Distance $$OA=\sqrt{2}$$

Hi Bunuel,

Is there any shorter way to solve this problem. Also, can you pls explain the logic behind the Skytail's approach in one of the above posts.
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29 Aug 2017, 07:24
this is a very very lengthy problem!
It's great for practice though!
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Joined: 18 Dec 2015
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09 Nov 2017, 16:57
Quicker way to solve this:

The midpoint of the diagonal is (3,4). The vector from the right bottom corner to the midpoint is V1 = <-3, -2>. You need to find a vector perpendicular to this vector and of the same magnitude and add it to the midpoint position vector to find the corners.

Vectors perpendicular meet:

-3x - 2y = 0 or y = (3/2) x
So the form is <x, 3/2x>

The two perpendicular vectors that have the same magnitude as V1 will meet:

x^2 + [(3/2)x]^2 = 13
x = +/- 2

So the two vectors are <2,3> or <-2,-3>

To find the corner closest to the origin add <-2,-3> to the position vector of the midpoint <3,4> to get <1,1>.

The magnitude of this vector sqrt(2) is the answer.
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08 Dec 2017, 03:57
1
akhil911 wrote:
Hi

Is there any shorter way to solve this problem.
I have understood the solution but this seems to be a very tough and complex problem.
I am wondering if this can be solved in some other way.

$$m = \frac{(y2 - y1)}{(x2-x1)}$$

$$m = \frac{6-2}{0-6} = \frac{-2}{3}$$

The other diagonal of a square is perpendicular to this one and we know slope of the perpendicular line is always $$\frac{-1}{m} = \frac{-1}{-2/3} = \frac{3}{2}$$

Now mid point of two coordinates or the diagonal of the square is $$(3,4)$$.

$$Slope = \frac{rise}{run}$$

From mid point of $$(3,4)$$, we need to back traverse a little. Since line is rising 3 and running 2, we reduce rise by 3 and run by 2 while back tracking to come up with the point $$(3-1,4-3)$$ i.e. $$(1,1)$$ which is the vertex of the square nearest to $$(0,0)$$

$$(0,0)$$ and $$(1,1)$$ be solved using Pythagoras theorem. $$\sqrt{1^2 + 1^2}$$ and this comes out to be $$\sqrt{2}$$
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15 Feb 2018, 08:56
I approached this problem more visually. I drew the two points, forming a diagonal. Then knowing that this is a square, I know that the intersection of the two diagonals must be perpendicular. I find the slope of the diagonal given (2/3) so I flip and make this negative to find the slope of the diagonal (-3/2). I used the midpoint formula to find the center of the square, or the diagonal intersection. That value is (3,4). So from 3,4 I move down 3 and over 2. That is the end of my other diagonal - I can't go further or else it won't be a square. I now have my figure and four endpoints.

Without doing any math, I know that the distance from 0,0 to 1,1 is root(2). I think leveraging the grid and drawing out the exact coordinates really helps with this problem.
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22 May 2018, 22:34
Bunuel wrote:
If 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. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

Responding to a pm:

Here:

http://www.veritasprep.com/blog/2013/04 ... ry-part-i/
This post will explain you the relevant concept.

and this post: https://www.veritasprep.com/blog/2013/0 ... y-part-ii/
discusses this question in detail.
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