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 . . .