Find a point on the y-axis that is equidistant from the points (6, −6) and (3, 3).

Question

Find a point on the y-axis that is equidistant from the points
(6, −6)
and
(3, 3).

Answer 1

The set of points equidistant from two given points is the perpendicular bisector of the line segment with the given points as its endpoints.  

 

Let A = (6,-6) and B = (3,3)

 

Midpoint of AB = (9/2, -3/2)

 

Slope of AB = 9/(-3) = -3

 

Perpendicular bisector has slope 1/3 and contains the point (9/2, -3/2)

 

Equation of perpendicular bisector:  y - (-3/2) = 1/3(x - 9/2)

 

                                                    y + 3/2 = (1/3)x - 3/2

 

                                                            y = (1/3)x - 3  (Has y-intercept (0,-3))

 

(0,-3) is on the perpendicular bisector of AB and is on the y-axis.  

 

Answer 2

Along the y-axis, x = 0. We use the distance formula, but we can skip the radical. (6 - 0)^2 + (-6 - y)^2 = (3 - 0)^2 + (3 - y)^2 36 + 36 + 12y + y^2 = 9 + 9 - 6y + y^2 y^2 + 12y + 72 = y^2 - 6y + 18 18y = -54 y = -3 (0,-3)