Question

find te point on the y-axis which is equidistan from (2,3) and (-4,1)​

Answer 1

Answer:

let point is A (0,y)

it is equidistant from C(2,3) andB(-4,1)

so AB=AC

AB= √[0-(-4)]^2+[y-1]^2

squaring it

(AB)^2= 16+y^2+1-2y

(AB)^2=17+y^2-2y

AC=√[0-2]^2+[y-3]^2

squaring it

(AC)^2=4+y^2+9-6y

(AC)^2=13+y^2-6y

we know AB= AC

so square of both are also same

then,

17+y^2-2y=13+y^2-6y

17-13=2y-6y

4=-4y

y=-1

so point is (0,-1)