If the point is equidistant from the points and , find k. Also, find the length of AP.

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.