Here's A Tough One...

**AlphaRock** · May 2nd 2008, 09:04 PM

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.

**o_O** · May 2nd 2008, 10:05 PM

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

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

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

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

etc. etc.

**Soroban** · May 3rd 2008, 07:18 PM

Hello, AlphaRock!

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

Point $P$ lies on the x-axis.
There are two points: . $A(0, 6)\,\text{ and }\,(0, -15)$
The sum of the distances of $PA$ and $PB$ is 27 units.
Determine the possible coordinates of $P.$

Code:

              A
              *
            / | \
          /   |6  \
        /     |     \
     Q* - - - + - - - *P
       \     O|   x  /
        \     |     /
         \    |15  /
          \   |   /
           \  |  /
            \ | /
             \|/
              *
              B

We have a kite-shaped quadrilateral $APBQ$
Its diagonals intersect at $O.$

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

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

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

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

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

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

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