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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