1. Here's A Tough One...

I know there are a lot of smart people on here.

Can anybody explain to me how to get this...?

Point P lines on the x axis. The coordinates of A and B. are (0, 6) and (0, -15) respectively. The sum of the distances of PA and PB is 27 units. Determine the possible coordinates of P.

I got (8, 0) (-8, 0), but did it with trial and error.

Respect to those who can get this.

2. Label point P: $\displaystyle (x_{0}, 0)$ (as it lies on the x-axis).

Making use of the distance formula between two points:
$\displaystyle PA = \sqrt{(x_{0} - 0)^{2} + (0 - (6))^{2}} = \sqrt{x_{0}^{2} + 36}$
$\displaystyle PB = \sqrt{(x_{0} - 0)^{2} + (0 - (-15))^{2}} = \sqrt{x_{0}^{2} + 225}$

You are given that: $\displaystyle PA + PB = 27$. So simply solve for x:

$\displaystyle \sqrt{x_{0}^{2} + 36} + \sqrt{x_{0}^{2} + 225} = 27$
$\displaystyle \sqrt{x_{0}^{2} + 225} = 27 - \sqrt{x_{0}^{2} + 36}$ (Moved one of the radicals to one side)
$\displaystyle x_{0}^{2} + 225 = \left(27 - \sqrt{x_{0}^{2} + 36}\right)^{2}$ (Squared both sides)
$\displaystyle x_{0}^{2} + 225 = 729 - 54\sqrt{x_{0}^{2} + 36} + x_{0}^{2} + 36$ (Expanded)
$\displaystyle 54\sqrt{x_{0}^{2} + 36} = 540$ (Simplified)
$\displaystyle \sqrt{x_{0}^{2} + 36} = 10$ (Square again)

etc. etc.

3. Hello, AlphaRock!

Another approach . . .
. . but the math is identical to O_o's solution.

Point $\displaystyle P$ lies on the x-axis.
There are two points: .$\displaystyle A(0, 6)\,\text{ and }\,(0, -15)$
The sum of the distances of $\displaystyle PA$ and $\displaystyle PB$ is 27 units.
Determine the possible coordinates of $\displaystyle P.$
Code:
              A
*
/ | \
/   |6  \
/     |     \
Q* - - - + - - - *P
\     O|   x  /
\     |     /
\    |15  /
\   |   /
\  |  /
\ | /
\|/
*
B
We have a kite-shaped quadrilateral $\displaystyle APBQ$
Its diagonals intersect at $\displaystyle O.$

We are given: .$\displaystyle AO = 6,\;BO = 15$
Let $\displaystyle PO = x$

We want: . $\displaystyle PA + PB \:=\:27$

In right triangle $\displaystyle AOP\!:\;\;x^2 + 6^2 \:=\:PA^2\quad\Rightarrow\quad PA \:=\:\sqrt{x^2+36}$

In right triangle $\displaystyle BOP\!:\;\;x^2 + 15^2 \:=\:PB^2\quad\Rightarrow\quad PB \:=\:\sqrt{x^2 + 225}$

Since $\displaystyle PA+ PB \:=\:27$, we have: .$\displaystyle \sqrt{x^2+36} + \sqrt{x^2+225} \:=\:27$

Then we can solve for $\displaystyle x$ as O_o suggested . . .