# 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 = 121+36 +12y + y^2

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

==> 26y = -99

==> y= -99/26

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

The distance d of (x1,y1) and(x2,y2) is given by:

d^2 =(x2-x)^2 + (y2+y1)^2.

Since PR = QR and R is on Y axis, the x coordinate of R is 0 and so the the coordinates of R are of the form (0,k)

Therefore PR= QR => PR^2* QR^2. Or

(0-3)^2+(k-7)^2 = (0-11)^2+(k- - 6)^2

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

-(14+12)k = 121+36 -9-49 = 99

k = 99/-26 = -(99/26)

The the coordinates of R = (0, -99/26)