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 (x_{1}, y_{1}) and (x_{2}, y_{2}) =

So, PA = QA

⇒√145= √(160 + y^{2} + 8y)

On squaring both the sides,

We get;

145 – 160 + y^{2} + 8y

y^{2} + 8y + 160 - 145 = 0

y^{2} + 8y + 15 = 0

y^{2} + 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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