Q. 84.9( 8 Votes )

# <span lang="EN-US

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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