Question

Find the point on the y-axis that is equidistant from (6,1) (-2,-3)

Answer 1

$Any \ point \ on \ the \ y axis \ can \ be \ stated \ as \ C(0,y) \\ \\ Distance \ Formula:\\\\ Given \ the \ two \ points \ (x _{1}, y _{1}) and (x _{2}, y _{2}), \\ \\the \ distance \ between \ these \ points \ is \ given \ by \ the \ formula: \\ \\ d= \sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2} \\ \\CA= \sqrt{(-2-0)^2 +(-3-y)^2}=\sqrt{(-2 )^2 +(-3-y)^2}=\sqrt{4 +9+6y+y^2 }=\\\\-\sqrt{ y^2+6y+13 }$

$CB= \sqrt{(1-y)^2 +(6-0)^2} =\sqrt{ 1-2y +y^2 +36} =\sqrt{ y^2-2y +37} \\ \\CA = CB \\ \\\sqrt{ y^2+6y+13 } =\sqrt{ y^2-2y +37} \ \ |^2\\\\ y^2+6y+13 = y^2-2y +37 \\ \\y^2+6y- y^2+2y =37-13\\ \\8y = 24 \ \ / :8 \\ \\y= 3 \\ \\C=(0,3)$

Answer 2

Looking the centre of the circle on the y-axis passing through the points (6;1) and  (-2,-3).
The coordinates of the center of this a circle are equal y-intercept of line segment bisector.

$(x-6)^2+(y-1)^2=(x+2)^2+(y+3)^2\\\\x^2-12x+36+y^2-2y+1=x^2+4x+4+y^2+6y+9\\-12x-2y+37=4x+6y+13\\-2y-6y=4x+12x+13-37\\-8y=16x-24\ \ \ \ \ |:(-8)\\y=-2x+3\\\\Answer:(0;\ 3)$