Thread: Define the set of points such that PA = NPB

1. 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

2. 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 $PA$ you mean the length of the line segment
$\overline{PA}$.
If $P: (r,s)$ then $(x-r)^2+(y-s)^2=N^2\cdot~[?]$

3. yes by PA I mean the distance from P to A

4. 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: . $P\!A \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB + PB \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB \:=\:n\!\cdot\!PB - PB$

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

$\Delta BDA \sim \Delta PFB$

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

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

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