# 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.

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