If the point <spa

Given, PA = PB

Squaring on both sides, we get

PA2 = PB2

We know that the distance between P(x1, y1 ) and Q(x2, y2 ) is .

(-2 – 2)2 + (k – 2)2 = (-2k – 2)2 + (-3 – 2)2

16 + k2 + 4 – 4k = 4k2 + 4 + 8k + 25

16 + k2 + 4 – 4k - 4k2 - 4 - 8k – 25 = 0

-3k2 – 12k – 9 = 0

Dividing by -3, we get

k2 + 4k + 3 = 0

By factorization method,

k2 + k + 3k + 3 = 0

k(k + 1) + 3(k + 1) = 0

(k + 1) (k + 3) = 0

k + 1 = 0 (or) k + 3 = 0

k = -1 (or) k = -3

Now, AP =

AP =

Case 1: When k = -1

AP =

=

√25

AP = 5 units

Case 2: When k = -3

AP =

=

AP = √41 units

The value of k is -1 (or) -3 and the length of AP is 5 units (or) √41 units respectively.

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