Sunday, January 3, 2016

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.

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)

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