Question

if a point A(0,2) is equidistant from the point B(3,p) and C(p,5),then find the value of p

Answer 1

given that the point a (0,2) uis equidistant from the point b (3,p) and the point c ( p,5)

as per the question 
ap² = ac²
ap = ac

by the distance formula

d = √(x2-x1)²+(y2-y1)²
here we have ab = ac
so the root gets cancelled

(3-0)² + (p-2)² = (p-0)² + (5-2)²
 9       + (p-2)² = (p)²    +  9
nine gets cancelled
we know by the formula that (a-b)² = a²+b²-2ab

therefore
p²-4p +4 = p²
p² gets cancelled on both the sides
-4p + 4 = 0
    p      = 4/4

therefore the value of p = 1

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Answer 2

Hiii friend,

A(0,2) is equidistant from the points B(3,P) and C(P,5)


Therefore,


AB = AC => AB² = AC²


A(0,2) and B(3,P)

Here,

X1 = 0 , Y1 = 2 and X2 = 3 , Y2 = P

AB² = (X2-X1)² + (Y2-Y1)² => ✓(3-0)² + (P-2)²

AB² = (3)² + (P)² + (2)² - 2 × P × 2 = 9+P²+4-4P

AB² = 13+P²-4P


And,


A(0,2) and C(P,5)


Here,


X1 = 0 , Y1 = 2 and X2 = P , Y2 = 5


AC² = (X2-X1)² + (Y2-Y1)² => (P-0)² + (5-2)²

AC² = (P)² + (3)²

AC² = P²+9


Now,


AB² = AC²

13+P² -4P = P²+9


P²-P² - 4P = 9-13

-4P = -4

P = -4/-4 => 1


Hence,


P = 1


HOPE IT WILL HELP YOU...... :-)