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Should be B. x^2= 2^2 + 6^2
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agdimple333

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:
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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agdimple333

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:
qimage040.gif

Bunuel Please shift the question in Problem Solving thread. This seems to be here by mistake.
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GMATinsight
agdimple333

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:
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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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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loser
Should be B. x^2= 2^2 + 6^2


Need detail explanation.

Thanks
Jahid
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Can someone explain the difference between what is referred to here as a perpendicular bisector and what I thought was just to draw a line segment from point J to y=4 with the point (6,4) being point K? I assumed the bisector was just to y=4, so how should I tell going forward for similar questions that the line needs to be extended another 6 units past y=4?
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