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In the diagram above, the line y = 4 is the perpendicular bisector of

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In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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In the diagram above, the line y = 4 is the perpendicular bisector of segment JK (not shown). What is the distance from the origin to point K ?

A. 4
B. \(2\sqrt{10}\)
C. 8
D. \(6\sqrt{2}\)
E. \(4\sqrt{34}\)

[Reveal] Spoiler:
Attachment:
qimage040.gif
qimage040.gif [ 1.15 KiB | Viewed 2768 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 May 2015, 08:41, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 26 May 2011, 00:25
Refer to diagram:
Attachment:
SOLN.JPG
SOLN.JPG [ 13.1 KiB | Viewed 4075 times ]

y=4 is perpendicular bisector of JK
Hence distance between Point J and line y = 6 = distance between line y and point K.
Point K(6,-2)
Distance between point K and origin is 6^2+2^2= 2\sqrt{10}

OA B.

Please let me know the OA.
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 26 May 2011, 00:40
coordinates of point k are (6,-2).
distance of J = 6 from y=4 means distance of 6 from y=4 for K gives y = -2.

thus 36+4 = distance from origin ^ 2
40 ^(1/2) = B
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 03 Nov 2011, 13:15
Should be B. x^2= 2^2 + 6^2
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 08 May 2015, 08:41
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 15 Mar 2016, 00:24
agdimple333 wrote:
Image
In the diagram above, the line y = 4 is the perpendicular bisector of segment JK (not shown). What is the distance from the origin to point K ?

A. 4
B. \(2\sqrt{10}\)
C. 8
D. \(6\sqrt{2}\)
E. \(4\sqrt{34}\)

[Reveal] Spoiler:
Attachment:
qimage040.gif

The point k is (6, -2)
Therefore, OK is root of 6^2 + 2^2 = root 40 or 2*root10
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 11 Jun 2016, 06:17
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agdimple333 wrote:
Image
In the diagram above, the line y = 4 is the perpendicular bisector of segment JK (not shown). What is the distance from the origin to point K ?

A. 4
B. \(2\sqrt{10}\)
C. 8
D. \(6\sqrt{2}\)
E. \(4\sqrt{34}\)

[Reveal] Spoiler:
Attachment:
qimage040.gif


Bunuel Please shift the question in Problem Solving thread. This seems to be here by mistake.
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 12 Jun 2016, 05:36
GMATinsight wrote:
agdimple333 wrote:
Image
In the diagram above, the line y = 4 is the perpendicular bisector of segment JK (not shown). What is the distance from the origin to point K ?

A. 4
B. \(2\sqrt{10}\)
C. 8
D. \(6\sqrt{2}\)
E. \(4\sqrt{34}\)

[Reveal] Spoiler:
Attachment:
qimage040.gif


Bunuel Please shift the question in Problem Solving thread. This seems to be here by mistake.

______________
Done. Thank you.
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 14 Nov 2016, 09:00
Answer:B
1- We have J(6,10) and the line y = 4 is the perpendicular bisector of segment JK ⇒ Intersection point of the two lines is the midpoint M(6,4) ⇒ k is (6,-2)
2- Taking the point on the X-axis L(6,0) ⇒ we have a right triangle OLK with O(0,0), L(6,0), and K(6,-2)
Finally the hypotenuse OK^2=OL^2+LA^2=6^2+2^2=36+4=40 ⇒ OK=sqrt(40)=2sqrt(10)
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Re: In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]

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New post 22 Feb 2017, 07:54
loser wrote:
Should be B. x^2= 2^2 + 6^2



Need detail explanation.

Thanks
Jahid
Re: In the diagram above, the line y = 4 is the perpendicular bisector of   [#permalink] 22 Feb 2017, 07:54
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In the diagram above, the line y = 4 is the perpendicular bisector of

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