Maths - In Center

Equation of AB：x + 2y - 8 = 0Equation of AC：x - 2y + 8 = 0Equation pf BC：3x + 2y - 24 = 0Let O(a, b) be the in-center, as AO is the angle bisector of ∠BAC, so, b = 4.The distance between O and AC is |a - 2(4) + 8|/√5and, the distance between O and BC is |3a + 2(4) - 24|/√13So, |a - 2(4) + 8|/√5 = |3a + 2(4) - 24|/√13==> a²/5 = (9a² - 96a + 256)/13==> 13a² = 45a² - 480a + 1280==> 32a² - 480a + 1280 = 0==> a² - 15a + 40 = 0==> a = [15 ± √(15² - 4*40)]/2==> a = (15 + √65)/2 (rej) or (15 - √65)/2
∴ the co-ordinates of the in-center is ((15 - √65)/2, 4)

AB=[(x-8)^2 +(4-0)^2]^0.5
AB=80^0.5
BC=[(8-4)^2 +(0-6)^2]^0.5
BC=52^0.5
AC=[(4-0)^2 +(6-4)^2]^0.5
AB^2=AC^2 +BC^2-2(AC)(BC)cos ACB
cos ACB=-65^-0.5
AC^2=AB^2+BC^2-2(AB)(BC)cos ABC
cos ABC=7(65)^-0.5
tan ABO=(0-8)/(4-0)
tan ABO=-2
If P point at PB line and make CBP=ABP
m of PB=tan(180+ ABO-0.5ABC)
m of PB=-4.831955546
y=(-4.831955546)x+c
0=(-4.831955546)(8)+c
c=38.65564437
y=(-4.83195546)x+(38.65564437) i
If E at x-axis and make ECB=ECA
CEB=180+ABO-ABC-0.5ACB
CEB=38.25766171
m of EC=tan CEB
m of EC=0.788553284
y=(0.788553284)x+c
6=(0.788553284)(4+c
c=2.845786863
y=(0.788553284)x+(2.845786863) ii
The co-ordinates of its in center is (x,y)=(7.38392777,2.97683363)

Method 1：
http://s22.postimg.org/4pkobqaz4/image.jpg

Method 2：
http://s28.postimg.org/kzrxsgyuz/image.png

2015-07-01 00:10:48 補充：
If you don't understand |ax+by+c|/√(a²+b²),
it may help you
http://s10.postimg.org/bu7tlavtz/image.png