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# On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia

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On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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15 Oct 2016, 06:49
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95% (hard)

Question Stats:

58% (02:21) correct 42% (02:05) wrong based on 213 sessions

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On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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15 Oct 2016, 22:36
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Manonamission wrote:
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)

Checking option after having understood several things seems the best way forward in this Questions

Check solution in attachment

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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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15 Oct 2016, 22:08
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Length of the diagonal = sqrt(36 + 16) = sqrt(52)
sqrt(2) * Side = sqrt(52)
Length of each side = sqrt(26)

Test the length in the given options, we find that only option C satisfies the length of the side of the square.

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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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15 Oct 2016, 23:53
Vyshak wrote:
Length of the diagonal = sqrt(36 + 16) = sqrt(52)
sqrt(2) * Side = sqrt(52)
Length of each side = sqrt(26)

Test the length in the given options, we find that only option C satisfies the length of the side of the square.

can you explain in detail please!!
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On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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16 Oct 2016, 00:01
Manonamission wrote:
Vyshak wrote:
Length of the diagonal = sqrt(36 + 16) = sqrt(52)
sqrt(2) * Side = sqrt(52)
Length of each side = sqrt(26)

Test the length in the given options, we find that only option C satisfies the length of the side of the square.

can you explain in detail please!!

Question asks about the corners of the square. A line joining the corner of the diagonal to an adjacent corner must be the side of the square.

So the options given must first satisfy the criteria of being a corner of the given square. Only option C satisfies the criteria.
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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27 Oct 2016, 08:24
Manonamission wrote:
Vyshak wrote:
Length of the diagonal = sqrt(36 + 16) = sqrt(52)
sqrt(2) * Side = sqrt(52)
Length of each side = sqrt(26)

Test the length in the given options, we find that only option C satisfies the length of the side of the square.

can you explain in detail please!!

I used slightly different approach, but it seems like the one Vyshak meant to describe.

1. Draw the diagonal.
2. You see that point (0;2) cant be the corner, otherwise it will be not a square. So the corner cant have coordinates like (0;y) or (x:0). So only answers C and E is out.

If you can not see it right away then after drawing the diaganal, go to point 3 right away:
3. tTry answers and see whether the point can give you a squire
draw point and connect corners of our squire. you see that one side from the left is 5, but the other side from the bottom is not 5 clearly.
The same case with answers B D E

If you need detailed explanation about answers pls let me know. I would be glad to help
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On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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Updated on: 17 Jun 2017, 03:25
Manonamission wrote:
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)

This question is possibly designed to mislead test takers into thinking that sides of this square must be perfectly straight. Actually, we can also have a tilted square. The most natural solution would be to think that (0,2) must be the closet point to the origin however that is actually incorrect- it's nearly optical illusion- mathematically it could not be (0,2) because then one of the side lengths would be 4 (6-2 =4) and the other side length would be 6. This square is actually tilted- but more to the point, this question seems to loosely imply that these two endpoints that form a diagonal are also vertices. If that is true- then the point closest to the origin must be equidistant (equally distant) from both of those coordinates. Apply the distance formula

Thus
"C"
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Originally posted by Nunuboy1994 on 17 Jun 2017, 03:00.
Last edited by Nunuboy1994 on 17 Jun 2017, 03:25, edited 1 time in total.
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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17 Jun 2017, 03:03
Manonamission wrote:
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)

Bunuel it seems somewhat reasonable that this question is suggesting the two endpoints of the diagonal of the square must also be vertices- in which case the point that is closest to the origin must be equally distant. However, can we not actually have a diagonal that cuts through a square but does not necessarily connect two vertices? Kudos for explanation.
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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17 Jun 2017, 04:54
Nunuboy1994 wrote:
Manonamission wrote:
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What are the coordinates of the point on the corner of the square which is closest to the origin?

(A) (0, 1)
(B) (1, 0)
(C) (1, 1)
(D) (2, 0)
(E) (2, 2)

Bunuel it seems somewhat reasonable that this question is suggesting the two endpoints of the diagonal of the square must also be vertices- in which case the point that is closest to the origin must be equally distant. However, can we not actually have a diagonal that cuts through a square but does not necessarily connect two vertices? Kudos for explanation.

Not sure I can follow you...

The endpoints of the diagonal of a square and vertices of a square are the same thing - 2 diagonals, each has two endpoints = 4 vertices.
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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14 Aug 2017, 04:59
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Clearly C

Just put the values to get slopes m1 and m2, andwe know m1 x m2 =-1
lets us put values for option C for m1 and m2
m1=(6-1)/(0-1) = -5
m2= (1-2)/(1-6) = 5
clearly m1 x m2 = -1
since we are multiplying slopes of side of square... these values are satisfying here so it is the shortest way.
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia  [#permalink]

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26 Nov 2017, 06:32
Distance between co-ordinates (0,6) and coordinate we are going to find out = Distance between co-ordinate (6,2) and co-ordinate we are going to find out.

Only C satisfies this: \sqrt{$$(0-1)^2 + (6-1)^2$$} = \sqrt{$$(1-6)^2 + (1-2)^2$$}
Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints of the dia &nbs [#permalink] 26 Nov 2017, 06:32
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