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

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P(3,7)  Q(11,-6)

We have the point R on the y-axis,

==> R (0,y)

Such that:

PR = QR

PR= sqrt[(3-0)^2 + (7-y)^2]

QR= sqrt[(11-0)^2 + (-6-y)^2

==> sqrt[(9+(7-y)^2]= sqrt[121 + (-6-y)^2]

Square both sides:

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

Open brackets:

==> 9 + 49 -14y + y^2...

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P(3,7)  Q(11,-6)

We have the point R on the y-axis,

==> R (0,y)

Such that:

PR = QR

PR= sqrt[(3-0)^2 + (7-y)^2]

QR= sqrt[(11-0)^2 + (-6-y)^2

==> sqrt[(9+(7-y)^2]= sqrt[121 + (-6-y)^2]

Square both sides:

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

Open brackets:

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

==> 58 -14y = 157 + 12y

==> 26y = -99

==> y= -99/26

==> R (0, -99/26)

Approved by eNotes Editorial Team