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

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18 Sep 2009, 12:21
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You don't need to use the distance formula here; instead we can use slopes. I posted this solution a while back to another forum:

We have two endpoints of a diagonal of a square. We can use the following:

-the midpoint of one diagonal is the midpoint of the other diagonal;
-the diagonals are perpendicular.

If (0,6) and (6,2) are endpoints, (3,4) is the midpoint.

From (0,6) to (3,4), we go right 3 and down 2; that is, we increase x by 3 and decrease y by 2: the slope is -2/3. Consider the perpendicular diagonal- its slope is the negative reciprocal, i.e. 3/2. From (3,4), on a perpendicular line, to find a point the same distance from (3,4) as (0,6) is, we can decrease x by 2 and decrease y by 3, or we can increase x by 2 and increase y by 3. The endpoints of the other diagonal are (1,1) and (5,7).

The distance from the origin to (1,1) is sqrt(2).
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05 Nov 2009, 13:23
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Won't the square will be as shown with red lines ,given that 6,2 and 0,6 are end points of diagonals ?

so nearest point on vertex to square must be (0,2)

dist from 0,0 to 0,2 is 2

where am I going wrong ?
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nearestverexonsquare.JPG [ 7.39 KiB | Viewed 5072 times ]

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25 Nov 2009, 04:33
powerka wrote:
I spent an exorbitant amount of time on this problem.

Tried to tackled it via algebra, and got stuck.

This is what I did:
. midpoint= (3,4)
. slope= -2/3
. then perpendicular slope= 2/3
. plugging in (3,4) into y=3/2x+k got k=-1/2 -> y=3/2x-1/2 .....(eq1)
. distance between points (diagonal)= sqrt(6^2+4^2)= sqrt(52)
. half the distance= sqrt(13)
. then distance between midpoint and vertex is given by: sqrt[(x-3)^2+(y-4)^2]=sqrt(13)->(x-3)^2+(y-4)^2=13 ....(eq2)
. plugging equation1 into equation2 I tried to solve to get vertex (1,1) as said by other posters
. but did not succeed. Reached x^2-3x+5=0, which has no real roots.

Good news is that googling around I found a fantastic shortcut:

From (3,4), on a perpendicular line (with slope 3/2), to find a point the same distance from (3,4) as (0,6) is, we decrease x by 2 and decrease y by 3, thus getting vertex (1,1), or we increase x by 2 and increase y by 3, thus getting vertex (5,7).

The distance from (0,0) to (1,1) is sqrt(2).

u mean perpendicular slope= 3/2 right?

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25 Feb 2010, 09:32
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Pedros 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)

Ans:
[Reveal] Spoiler:
C

Given endpoints of diagonal of a square: B(0,6) and D(6,2). Let other vertices be A (closest to the origin, left bottom vertex) 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: $$M(x,y)=(\frac{6+0}{2},\frac{2+6}{2})=(3,4)$$.

Slope of the line segment AM*Slope of the line segment BD=-1 (as they are perpendicular to each other) --> $$\frac{y-4}{x-3}*\frac{6-2}{0-6}=-1$$ --> $$\frac{y-4}{x-3}=\frac{3}{2}$$ --> $$y-4=\frac{3}{2}(x-3)$$

Distance between the unknown vertices to the midpoint is half the diagonal:
$$(x-3)^2+(y-4)^2=(\frac{\sqrt{52}}{2})^2=13$$ --> $$(x-3)^2+\frac{9}{4}(x-3)^2=13$$ --> $$(x-3)^2=4$$ --> $$x=1$$ or $$x=5$$ --> $$y=1$$ or $$y=7$$

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

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02 Mar 2010, 02:33
Thanks for the detailed explanation.

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03 Aug 2010, 05:31
Hi,

The way I see this is that the vertexes are (0,6), (6,6), (0,2) and (6,2).
The closest from (0,0) is (0,2) and hence the distance is sqrt (0^2+2^2) =2

What am I missing here?

regards,
Jack

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03 Aug 2010, 14:21
jakolik wrote:
Hi,

The way I see this is that the vertexes are (0,6), (6,6), (0,2) and (6,2).
The closest from (0,0) is (0,2) and hence the distance is sqrt (0^2+2^2) =2

What am I missing here?

That is not a square - it's a rectangle. Its length is 6 and its width is 4. In this question, the square is slightly rotated; it does not have sides parallel to the x and y axes.
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04 Aug 2010, 11:44
Hey Bunuel,

I understand how the square is placed and I have attached the same for everyone's benefit.

However I want your help for the below things:
(I have not been able to paste the formulae that you have typed)

(1) Length of the diagonal formula. Should we use that formula every time we want to derive the length of diagonal.

(2) The formula that you used for the "Distance between the unknown vertices to the midpoint is half the diagonal" in the very end.
Attachments

Graph.jpg [ 271.12 KiB | Viewed 5192 times ]

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04 Aug 2010, 11:52
seekmba wrote:
Hey Bunuel,

I understand how the square is placed and I have attached the same for everyone's benefit.

However I want your help for the below things:
(I have not been able to paste the formulae that you have typed)

(1) Length of the diagonal formula. Should we use that formula every time we want to derive the length of diagonal.

(2) The formula that you used for the "Distance between the unknown vertices to the midpoint is half the diagonal" in the very end.

Distance between point $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ can be found by the formula $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$.

So the length of the diagonal would be the distance between two points B(0,6) and D(6,2): $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{52}$$.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Hope it helps.
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05 Aug 2010, 02:18
OMG....completely missed all thse informations

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11 Sep 2010, 13:01
Square: ABCD,
Midpoint: M

A(0;6)
C(6;2)

Midpoint is between A and C: M(3;4)

To get the closest vertex:
1. move the square and make the origin the midpoint. M'(0;0), so C'(3;-2)
2. rotate C' clockwise 90 degrees around M' to get B (or D), so C''(-2;-3) = B(-2;-3)
3. move back the square to the original position, so B'(1;1)
4. calculate distance between B' and Origin: sqrt(2)*1

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11 Sep 2010, 22:38
nonameee wrote:
Is it a real GMAT question? I can't even write the solution under 2 minutes (not to mention to solve it under 2 minutes).

Yes it doesn't seem to be realistic GMAT question as it requires lengthy calculations and even backsolving can not help much to get the right answer quickly. But good for practice though.
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19 Sep 2010, 08:01
I found the following way, it seems faster but not perfect I guess.
1) Denote the coordinates of the point we are looking for as (m;n)
2) Build 2 equations based on the sides of the square: x^2=(6-n)^2+m^2 and x^2=(6-m)^2+(2-n)^2
3) Solving these equations we can get 3m-1=2n
4) We can assume that according to the draft m<6 and n<2
5) Thus, a possible solution is n=1, m=1

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21 Sep 2010, 17:37
thanks for the great explanation Bunuel.

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22 Sep 2010, 22:45
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please kudos me if it helps

now we know that diagonal as calculated is $$\sqrt{52}$$
we also know that diagonal of a square is a$$\sqrt{2}$$ , where a is the side
Therefore one side is $$\sqrt{26}$$.
Let C (x, y) be the closest vertice
now all sides are equal in a square
Therefore distance between C and (0,6) and C and (6,2) will be equal to$$\sqrt{26}$$

solve for x and y the following equations
$$(x-0)^2 + (y-6)^2 =26$$and
$$(x-6)^2 + (y-2)^2 = 26$$
we get C= (1,1)
hence the distance is $$\sqrt{2}$$ ANS

If it was a help Please kudos me

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23 Sep 2010, 11:18
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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).
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26 Oct 2010, 04:36
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

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26 Oct 2010, 06:30
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Beauty of Graphs - Example 2
+ Stunning Symmetry of Squares

If AC is the given diagonal with co-ordinates (0, 6) and (6, 2), we see it is a sloping line. The center point of the line is (3, 4) obtained by averaging x and y co-ordinates as given: (0 + 6)/2 = 3 and (6 + 2)/2 = 4.
Think of a horizontal line PQ whose y co-ordinates were shifted by 2 units up and 2 units down to obtain AC

Attachment:

Graph1.jpg [ 9.53 KiB | Viewed 4297 times ]

Diagonal BD will be perpendicular to AC and will pass through (3, 4). It will have a corresponding line RS whose x co-ordinates will be shifted by 2 units right and 2 units left to obtain the co-ordinates of BD. (This is so because y co-ordinates of PQ were shifted by 2 units to obtain AC and PQ is perpendicular to RS)

Attachment:

Graph2.jpg [ 12.98 KiB | Viewed 4295 times ]

The closest co-ordinate to (0, 0) is (1, 1) and its distance from center is $$\sqrt{2}$$
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26 Oct 2010, 06:49
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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04 Apr 2011, 23:39
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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# m15q19: (6, 2) and (0, 6) are the endpoints of the diagonal

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