If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find the value of p. Also, find the length of AB.

Given that point A is equidistant from points B and C, so AB = AC

By distance formula, as shown below:

AB = √{(3 – 0)² + (p – 2)²}

= √{(3)² + (p – 2)²}

= √{9 + p² – 4p + 4}

⇒ AB = √{13 + p² – 4p}

AC = √{(p – 0)² + (5 – 2)²}

= √{(p)² + (3)²}

⇒ AB = √{9 + p²}

Now, AB = AC

Squaring both sides, we get,

(13 + p² – 4p) = (9 + p²)

⇒ 4 = 4p

⇒ p = 1

Now, AB = √{13 + p² – 4p}

⇒ AB = √(13 + 1 – 4)

= √10 units

Therefore, the distance of AB = √10 units.