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Math Help - Define the set of points such that PA = NPB

  1. #1
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    Define the set of points such that PA = NPB

    Let us say you have two points in 2-D, point A and point B.

    A has position (a,b)
    B has position (c,d)

    Let N be a real number > 0

    Define the set of points such that the distance PA = N*PB
    Last edited by Corpsecreate; June 4th 2011 at 10:59 AM.
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  2. #2
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    Quote Originally Posted by Corpsecreate View Post
    Let us say you have two points in 2-D, point A and point B.
    A has position (x,y)
    B has position (u,v)
    Let N be a real number > 0
    Define the set of points such that the distance PA = N*PB
    I assume that by PA you mean the length of the line segment
    \overline{PA}.
    If P: (r,s) then (x-r)^2+(y-s)^2=N^2\cdot~[?]
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  3. #3
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    yes by PA I mean the distance from P to A
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  4. #4
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    Hello, Corpsecreate!

    \text{Let us say you have two points in 2-D.}
    \text{Point }A\text{ has position }(a,b).\;\text{ Point }B\text{ has position }(c,d).

    \text{Let }n\text{ be a real number }> 0 \;\hdots\;\text{and }n \ne 1.

    \text{Define the set of points such that: }\:\overline{P\!A} \:=\:n\!\cdot\!\overline{PB}
    Code:
                                        P
                                        o (x,y)
                                     *  *
                                  *     * y-d
                            B  *        *
                       (c,d)o  *  *  *  o F
                         *  *           *
                      *     *           *
                   *        * d-b       *
                *           *           *
             *              *           *
     (a,b)o  *  *  *  *  o  *  *  *  *  o
          A - - - c-a - - - D  - x-c -  E
          : - - - - - - x-a - - - - - - :

    We have: . P\!A \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB + PB \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB \:=\:n\!\cdot\!PB - PB

    . . . . . . . . AB \:=\:(n-1)PB \quad\Rightarrow\quad \frac{AB}{PB} \:=\:n-1


    \Delta BDA \sim \Delta PFB

    . . \frac{AD}{BF} \,=\,\frac{AB}{PB} \quad\Rightarrow\quad \frac{c-a}{x-c} \,=\,n-1 \quad\Rightarrow\quad \boxed{x \:=\:\frac{c-a}{n-1} + c}

    . . \frac{BD}{PF}\,=\,\frac{AB}{PB} \quad\Rightarrow\quad \frac{d-b}{y-d} \,=\,n-1 \quad\Rightarrow\quad \boxed{y \:=\:\frac{d-b}{n-1} + d}

    Is this what you're looking for?

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