# Math Help - balance point in a triangle

1. ## balance point in a triangle

Hey guys,
I have a work sheet to do and I did almost all the questions except 9.
http://i13.tinypic.com/4c7nz8h.jpg
I need you help how to do that.
I added all the answers from the previous questions in this sheet.

A(0,6) B(0,-2) C(12,2)
mid point A to B= (0,2)
mid point A to C= (6,4)
mid point B to C= (6,0)
Intersection point(4,2)

Thanks a lot!

2. Hello, dgolverk!

Your intersection point is incorrect . . .

$A(0,6),\;B(0,-2),\; C(12,2)$

$\begin{array}{ccc}\text{midpoint of }AB = D(0,2) \\ \text{midpoint of }BC = E(6,0) \\ \text{midpoint of }AC = F(6,4)\end{array}$

$\text{Intersection point: }P(4,2)$ . . . no

Since $AB$ is vertical and its midpoint is $D(0,2)$,
. . its perpendicular bisector is the horizontal line: $y \,= \,2$

The slope of $AC$ is: . $m \:=\:\frac{2 - 6}{12 - 0} \:=\:\frac{-4}{12}\:=\:-\frac{1}{3}$

The perpendicular slope is: . $M_{\perp} = +3$

The line through $F(6,4)$ with slope $3$ is:
. . $y - 4\:=\:3(x - 6)\quad\Rightarrow\quad y \:=\:3x-14$

The two perpendicular bisectors: $y \:=\:3x - 14$ and $y \,= \,2$
. . intersect when: . $3x - 14 \:=\:2\quad\Rightarrow\quad x = \frac{16}{3}$

Therefore, the intersection point is: . $\boxed{P\left(\frac{16}{3},\,2\right)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We must verify that $P$ is equidistant from $A,\,B,\,C.$

Distance formula
The distance between $(x_1,\,y_1)$ and $(x_2,\,y_2)$ is: . $d \;=\;\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$

Distance from $P\left(\frac{16}{3},\,2\right)$ to $A(0,6)$
. . $PA \:=\:\sqrt{\left(\frac{16}{3} - 0\right)^2 + (2 - 6)^2} \:=\:\sqrt{\frac{256}{9} + 16} \:=\:\sqrt{\frac{400}{9}} \:=\:\boxed{\frac{20}{3}}$

Distance from $P\left(\frac{16}{3},\,2\right)$ to $B(-,-2)$
. . $PB \:=\:\sqrt{\left(\frac{16}{3} - 0\right)^2 + \left(2 - [-2]\right)^2} \:=\:\sqrt{\frac{256}{9} + 16} \:=\:\sqrt{\frac{400}{9}} \:=\:\boxed{\frac{20}{3}}$

Distance from $P\left(\frac{16}{3},\,2\right)$ to $C(12,2)$
. . $PC \:=\:\sqrt{\left(\frac{16}{3} - 12\right)^2 + (2 - 2)^2} \;=\;\sqrt{\left(\frac{-20}{3}\right)^2 + 0^2} \:=\:\sqrt{\frac{400}{9}} \:=\:\boxed{\frac{20}{3}}$

. . . There!

3. Thanks a lot!