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In the diagram above, the line y = 4 is the perpendicular bisector of [#permalink]
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25 May 2011, 12:09
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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. \(4\sqrt{34}\) Attachment:
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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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26 May 2011, 00:25
Refer to diagram: Attachment:
SOLN.JPG [ 13.1 KiB  Viewed 4139 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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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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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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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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15 Mar 2016, 00:24
agdimple333 wrote: 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. \(4\sqrt{34}\) 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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11 Jun 2016, 06:17
agdimple333 wrote: 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. \(4\sqrt{34}\) 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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12 Jun 2016, 05:36



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