Please explain the following in easy to understand steps.Point P lies on the x-axis. The coordinates of A and B are (3,4) and (5,8). Determine the...

Since the point P lies on the x-axis, then the coordinate
is ( x, 0).

Now given A(3,4) and B(5,8) such
that:

The distance AP = the distance
BP

==> l AP l = l BP
l

Now we will use the distance
formula:

l AP l = sqrt[(xA- xP)^2 +
(yA-yP)^2]

         = sqrt(3 - x)^2 +(4
-0)^2

           = sqrt(9-6x+x^2 +
16)

           = sqrt(x^2 -6x +
25)...............(1)

l BP l = sqrt[(xB - xP)^2 + (yB -
yP)^2]

          = sqrt[(5-x)^2 +
(8-0)^2]

           = sqrt(25-10x+x^2 +
64)

            = sqrt(x^2 -10x +
89)..............(2)

We know
that:

l AP l = lBP
l

==> eq. (1) = eq.
(2)

==>sqrt(x^2 -6x + 25) = sqrt(x^2 - 10x +
89)

Square both
sides:

==>x^2 - 6x +25 = x^2 -10x +
89

==>-6x + 25 = -10x +
89

==> 4x =
89-25

==> 4x =
64

==>x=
16

Then the point P is  (16,
0)