# Thread: FInd the incenter

1. ## FInd the incenter

A triangle is located at A(-2,-5), B(0,7), and C(6,1)

Find the incenter.

I get the answer (4,3), but that's impossible. I'm stuck on this question for half an hour!

Thanks.

2. Originally Posted by chengbin
A triangle is located at A(-2,-5), B(0,7), and C(6,1)

Find the incenter.

I get the answer (4,3), but that's impossible. I'm stuck on this question for half an hour!

Thanks.

my approach
let incentre is at (x,y) , then it will be equidistant from A(-2,-5), B(0,7), and C(6,1)
using distance formula
$\displaystyle (x+2)^2+(y+5)^2=(x-0)^2+(y-7)^2$ ..........(1)
$\displaystyle (x+2)^2+(y+5)^2=(x-6)^2+(y-1)^2$ ..........(2)
solving (1)and (2),
x+6y=5 and 4x+3y=2

which gives

$\displaystyle x= \frac{-1}{7}$ and $\displaystyle y= \frac{6}{7}$

3. Using the formula $\displaystyle \frac {ax_1+bx_2+cx_3}{a+b+c}, \frac {ay_1+by_2+cy_3}{a+b+c}$

$\displaystyle a=6\sqrt 2, b=10, c=2\sqrt {37}$

$\displaystyle \frac {-12\sqrt 2 +12\sqrt {37}}{6\sqrt 2+10+2\sqrt {37}}, \frac{-30\sqrt 2+70+2\sqrt {37}} {6\sqrt 2+10+2\sqrt {37}}$

Oh I see my error, but I don't know how to get your answer with this formula.

Actually, your answer is wrong too. Your knowledge of the incenter is incorrect.

4. Originally Posted by ramiee2010

my approach
let incentre is at (x,y) , then it will be equidistant from A(-2,-5), B(0,7), and C(6,1)
using distance formula
$\displaystyle (x+2)^2+(y+5)^2=(x-0)^2+(y-7)^2$ ..........(1)
$\displaystyle (x+2)^2+(y+5)^2=(x-6)^2+(y-1)^2$ ..........(2)
solving (1)and (2),
x+6y=5 and 4x+3y=2

which gives

$\displaystyle x= \frac{-1}{7}$ and $\displaystyle y= \frac{6}{7}$
equidistant from the 3 vertices???

5. I got AB: y = 6x + 7 so its slope is +6
AC: y = 0.75x + 0.25 so its slope is +0.75
BC: y = -x + 7 so its slope is -1

I am not sure but I believe the slope of AO = (AB + AC)/2 = 27/8
then i am stuck at finding the slopes of BO, CO.
with their slopes found and the coordinates of 3 vertices given, the equations of AO, BO, CO can be found? and then the common solution (x, y) to these three be found???