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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) 1/sqrt(2)
b) 1
c) sqrt(2)
d) sqrt(3)
e) 2*sqrt(3)
if (0,6) and (6,2) are end points of a diagonal of a square - shouldnt they be two of the vertices of the square too?
also, if it were the vertices, the diagonal length shows that the quadrilateral isnt a square....which is where im most confused.... can anybody help?? thanks in advance.
this is what OA says:
First, we have to "guess" the coordinates of the vertex of the square closest to the origin. We can do it by making a sketch. The coordinates of the vertex are (1,1) and it is units away from the origin.
The correct answer is C.
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I cannot understand why do we have to do such complex calculations. If the vertices of the diagonal are 0,6 and 6,2 then the closest vertex will be (0,2). So the shortest distance from 0,0 should be 2.
Or may be I am missing something !
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Ans is C, \sqrt{2}for line 1 cross these two points: y=(-2/3)x+6 the middle point of line 1 is (3,4), and half of the diagonal is \sqrt{13}therefore line 2 is: y=(3/2)x-(1/2) suppose the vertex of line 2 is (x,y) then 13=(3-x)^2+(4-y)^2x=1, 5 so the closest vertex to (0,0) is (1,1) and the distance is \sqrt{2}
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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).
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Re: Difficult Geometry [#permalink]
 17 Sep 2009, 21:11
[/quote] Hi Yezz, It is a 2 variable equation with 2 equations. So it can be solved easily. a^2 + b^2 -12b + 36 = 26 a^2 + b^2 -12b-10=0 a^2=12b+10-b^2 (1*) ______________________________________________________ a^2 - 12a +36 + b^2 -4b +4 =26 a^2 - 12a + b^2 -4b +14=0 (putting 1*) 12b+10-b^2-12a+b^2-4b+14=0 8b-12a+24=0 2b-3a+6=0 a=(2b+6)/3 (2*) ____________________________________________ putting 2* in 1* You can solve.[/quote] Thanks Maliy, i appreciate. it was 3 am here and i tried to solve it for 15 minutes and i couldnt.  [/quote] How do you expect to do this much math in 2 mins. Gotta be a better way
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Re: Difficult Geometry [#permalink]
18 Sep 2009, 07:13
Some questions can be solved over 2 minutes. There are 75 minutes and 37 questions. Some can be done under 1 minute even under 30 seconds. Today I scored 51q and although 2 problems got my 6-7 minutes I finished exam in 64 minutes  That will give help you I think. So the easier under 1 that will give you a chance to solve hards in 5-6 minutes.
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Re: Difficult Geometry [#permalink]
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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Why to do these calculations and why we can not assume end points of diagonals as the vertices of square? Please explain flyingbunny wrote: Ans is C, \sqrt{2}
for line 1 cross these two points:
y=(-2/3)x+6
the middle point of line 1 is (3,4), and half of the diagonal is \sqrt{13}
therefore line 2 is:
y=(3/2)x-(1/2)
suppose the vertex of line 2 is (x,y)
then
13=(3-x)^2+(4-y)^2
x=1, 5
so the closest vertex to (0,0) is (1,1) and the distance is \sqrt{2}
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Re: coordinate geometry [#permalink]
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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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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Re: Coordinate plane [#permalink]
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: 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=7Hence the coordinates of the point A(1,1) and point C (5,7). Closest to the origin is A. Distance OA=\sqrt{2}Answer: C.
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Re: Coordinate plane [#permalink]
02 Mar 2010, 02:33
Thanks for the detailed explanation.
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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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Exactly, tht's what i did. But, the OE is way diffn. Just waiting for someone to come up with that explanation
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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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Re: Coordinate plane [#permalink]
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.
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Re: Coordinate plane [#permalink]
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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Re: Coordinate plane [#permalink]
05 Aug 2010, 02:18
OMG....completely missed all thse informations
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Re: Coordinate plane [#permalink]
01 Sep 2010, 01:36
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).
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Re: Coordinate plane [#permalink]
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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Re: Coordinate plane
[#permalink]
11 Sep 2010, 13:01
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