equation of points

**DINOCALC09** · December 2nd 2007, 03:21 PM

Find the equation of all points, P(x,y), which are aquidistant from the points A(-3,6) and B(2,1).

Thanks

**topsquark** · December 2nd 2007, 04:31 PM

Originally Posted by DINOCALC09

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
$\sqrt{(x - -3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 6)^2}$

The distance between B and (x, y) is
$\sqrt{(x - 2)^2 + (y - 1)^2}$

We want (x, y) such that these are equal, so
$\sqrt{(x + 3)^2 + (y - 6)^2} = \sqrt{(x - 2)^2 + (y - 1)^2}$

I'll let you simplify it.

-Dan

**Soroban** · December 2nd 2007, 04:37 PM

Hello, DINOCALC09!

I bet you already know enough to solve this . . .

Find the equation of all points, $P(x,y)$ , which are equidistant
from the points $A(-3,6)$ and $B(2,1).$

You know the Distance Formula, right?

Distance $PA \:=\:\sqrt{(x+3)^2 + (y-6)^2}$

Distance $PB \:=\:\sqrt{(x-2)^2 + (y-1)^2}$

Since these distances are equal: . $\sqrt{(x+3)^2 + (y-6)^2} \;=\;\sqrt{(x-2)^2 + (y-1)^2}$

Square both sides: . $(x+3)^2 + (y-6)^2\;=\;(x-2)^2 + (y-1)^2$

Expand: . $x^2 + 6x + 9 + y^2 - 12y + 36 \;=\;x^2 - 4x + 4 + y^2 - 2y + 1$

And this simplifies to: . $10x - 10y \:=\:40\quad\Rightarrow\quad \boxed{y \:=\:x + 4}$

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