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
$(x+2)^2+(y+5)^2=(x-0)^2+(y-7)^2$ ..........(1)
$(x+2)^2+(y+5)^2=(x-6)^2+(y-1)^2$ ..........(2)
solving (1)and (2),
x+6y=5 and 4x+3y=2

which gives

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

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

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

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

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

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
$(x+2)^2+(y+5)^2=(x-0)^2+(y-7)^2$ ..........(1)
$(x+2)^2+(y+5)^2=(x-6)^2+(y-1)^2$ ..........(2)
solving (1)and (2),
x+6y=5 and 4x+3y=2

which gives

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