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= √((-2-0)^2 +(-3-y)^2)=√((-2 )^2 +(-3-y)^2)=√(4 +9+6y+y^2 )=\\\\-√( y^2+6y+13 )`


`CB= √((1-y)^2 +(6-0)^2) =√( 1-2y +y^2 +36) =√( y^2-2y +37) \\ \\CA = CB \\ \\√( y^2+6y+13 ) =√( 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)`