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Math Help - Here's A Tough One...

  1. #1
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    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.
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  2. #2
    o_O
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    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.
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  3. #3
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    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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