What is the value of x and y if the point (x, y) is equidistant from (3, 6) and (4,8)?

The distance d between the points (x1,y1) and (x2,y2) is given
by:

d^2 = (x2-x1)^2+(y2-y1)^2.

Given
that (x,y) is equidistant from the points (3,6) and (4,8), we have to find (x,y). But we  know by
geometry that  a lot of solutions will be there. Actually every point on the perpendicular
bisector of the line segment joining (3,6) and (4,8) is equidistant from (3,6) and
(4,8)

The distance between (x,y) and (3,6) is given
by:

d^2 = (3-x)^2+(6-y)^2....(1)

The
distance between (x,y) and the point (4,8) is given by:

d^2=
(4-x)^2+(8-y)^2....(2)

 The distances at (1) and (2) are equal by
data. Therefore ,

(3-x)^2+(6-y)^2 =
(4-x)^2+(8-y)^2.

9 -6x+x^2 +36-12y +y^2 = 16
-8x+x^2+64-16y+y^2.

45-6x-12y = 80 -8x -16y . Other terms
cancel.

8x-6x + 16y-12y = 80-45

2x+ 4y
= 35.

Therefore all the points on 2x+4y - 10 are equidistant from
(3,6) and (4,8).