# Define the set of points such that PA = NPB

• Jun 4th 2011, 09:42 AM
Corpsecreate
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
• Jun 4th 2011, 09:52 AM
Plato
Quote:

Originally Posted by Corpsecreate
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 $\displaystyle PA$ you mean the length of the line segment
$\displaystyle \overline{PA}$.
If $\displaystyle P: (r,s)$ then $\displaystyle (x-r)^2+(y-s)^2=N^2\cdot~[?]$
• Jun 4th 2011, 09:54 AM
Corpsecreate
yes by PA I mean the distance from P to A
• Jun 8th 2011, 04:56 AM
Soroban
Hello, Corpsecreate!

Quote:

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

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

$\displaystyle \text{Define the set of points such that: }\:\overline{P\!A} \:=\:n\!\cdot\!\overline{PB}$
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: .$\displaystyle P\!A \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB + PB \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB \:=\:n\!\cdot\!PB - PB$

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

$\displaystyle \Delta BDA \sim \Delta PFB$

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

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

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