Question

Find the point on the y axis which is equidistant from (2,-5) and

(-2,9)​

Answer 1

Answer:

(0,2)

Step-by-step explanation:

at y-axis x=0

using distance formula

√(X2-X1)^2 + (Y2-Y1)^2

(2-x)^2+(-5-y)^2=(-2-x)^2+(9-y)^2

4+x^2-4x+25+y^2+10y = 4+x^2+4x+81+y^2-18y

-8x-56+28y=0

since, x=0

28y=56

y=2