Find the equation of all points, P(x,y), which are aquidistant from the points A(-3,6) and B(2,1).
Thanks
Call one of the points that you are seeking (x, y). Then we will derive a relationship between x and y such that you can find all such points.
The distance between A and (x, y) is
$\displaystyle \sqrt{(x - -3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 6)^2}$
The distance between B and (x, y) is
$\displaystyle \sqrt{(x - 2)^2 + (y - 1)^2}$
We want (x, y) such that these are equal, so
$\displaystyle \sqrt{(x + 3)^2 + (y - 6)^2} = \sqrt{(x - 2)^2 + (y - 1)^2}$
I'll let you simplify it.
-Dan
Hello, DINOCALC09!
I bet you already know enough to solve this . . .
Find the equation of all points, $\displaystyle P(x,y)$, which are equidistant
from the points $\displaystyle A(-3,6)$ and $\displaystyle B(2,1).$
You know the Distance Formula, right?
Distance $\displaystyle PA \:=\:\sqrt{(x+3)^2 + (y-6)^2}$
Distance $\displaystyle PB \:=\:\sqrt{(x-2)^2 + (y-1)^2}$
Since these distances are equal: .$\displaystyle \sqrt{(x+3)^2 + (y-6)^2} \;=\;\sqrt{(x-2)^2 + (y-1)^2} $
Square both sides: .$\displaystyle (x+3)^2 + (y-6)^2\;=\;(x-2)^2 + (y-1)^2$
Expand: .$\displaystyle x^2 + 6x + 9 + y^2 - 12y + 36 \;=\;x^2 - 4x + 4 + y^2 - 2y + 1$
And this simplifies to: .$\displaystyle 10x - 10y \:=\:40\quad\Rightarrow\quad \boxed{y \:=\:x + 4}$