Posted by **Aldwin** on Wednesday, November 28, 2012 at 10:26am.

The coordinates of the end-points of a line segment PQ are P(3,7) and Q(11,-6). Find the coordinates of the point R on the y-axis such that PR = QR.

Please include the workings, as I would like to be able to perform these problems unassisted in the future (I just have trouble remembering the formulas).

Thank you anyone who answers this question :D

- Math -
**Steve**, Wednesday, November 28, 2012 at 10:39am
A point on the y-axis has coordinates (0,y).

The distance from (0,y) to P(3,7) is the diagonal of a rectangle which has sides (3-0) and (7-y)

similarly for Q: the distances are (11-0) and (-6-y)

So,

PR = √(3^2 + (7-y)^2)

QR = √(11^2 + (-6-y)^2)

to have PR=QR, you need

√(3^2 + (7-y)^2) = √(11^2 + (-6-y)^2)

square both sides to get rid of the radicals:

9+(7-y)^2 = 121+(6+y)^2

9+49-14y+y^2 = 121+36+12y+y^2

collect terms (the y^2's go away - yay!)

26y = 99

y = 99/26

so, the point is (0,99/26)

odd answer, but hey, 99/26 is just a number, like 3 or -5.

- Math -
**Steve**, Wednesday, November 28, 2012 at 10:41am
oops,

26y = -99, so

y = -99/26

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