Question 1060124
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The co ordinates 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
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The points in the y-axis are {R = (0,y)}.

The distance from R to P is  {{{sqrt((-3)^2  + (y-7)^2)}}} = {{{sqrt(9 + (y-7)^2)}}}.

The distance from R to Q is  {{{sqrt(11^2  + (y-(-6))^2)}}} = {{{sqrt(121+(y+6)^2)}}}.


The condition  |PR| = |QR|  is

{{{sqrt(9 + (y-7)^2)}}} = {{{sqrt(121+(y+6)^2)}}}.

Square both sides to get

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

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

-99 = 26y.

y = {{{-99/26}}}.
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