Q. 84.9( 8 Votes )

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Answer :

Given,

A (2, - 4) is equidistant from P (3, 8) = Q ( - 10, y) is equidistant from A (2, - 4)


By distance formula;


Distance between two points (x1, y1) and (x2, y2) =


So, PA = QA






145= (160 + y2 + 8y)


On squaring both the sides,


We get;


145 – 160 + y2 + 8y


y2 + 8y + 160 - 145 = 0


y2 + 8y + 15 = 0


y2 + 5y + 3y + 15 = 0


y(y + 5) + 3(y + 5) = 0


(y + 5)(y + 3) = 0


If y + 5 = 0, then y = - 5


If y + 3 = 0, then y = - 3


y = - 3, – 5


Now,


Distance P (3, 8) and Q ( - 10, y);



[Putting y = - 3]




Again, distance between P (3, 8) and Q (-10, y);



[Putting y = - 5]




Hence,


The values of y are - 3, - 5 and corresponding values of PQ are √290 and √338=13√2 respectively.


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