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.
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
2 answers
oops,
26y = -99, so
y = -99/26
26y = -99, so
y = -99/26
2 answers