Question

find a point on y axis which is equidistant from P (3,2) Q (-1,-5)​

Answer 1

Answer:

Let A(5,−2) and B(−3,2). Then,

PA=PB

PA2=PB2

(5−0)2+(−2−y)2=(−3−0)2+(2−y)2

25+4+y2+4y=9+4+y2−4y

8y=−16

y=−2