there is a formula for solving the radius r of the inscribed circle
#r = sqrt(((s-a)(s-b)(s-c))/s)" "#
where #s=#half the perimeter of the triangle
and #s=1/2*(a+b+c)#
Let #A(3, 3), B(4, 2), C(8, 9)#
so that
#a=#distance from B to C
#b=#distance from A to C
#c=#distance from A to B
#a=sqrt((x_B-x_C)^2+(y_B-y_C)^2)#
#a=sqrt((4-8)^2+(2-9)^2)#
#a=sqrt(16+49)#
#a=sqrt(65)#
#b=sqrt((x_A-x_C)^2+(y_A-y_C)^2)#
#b=sqrt((3-8)^2+(3-9)^2)#
#b=sqrt(25+36)#
#b=sqrt(61)#
#c=sqrt((3-4)^2+(3-2)^2)#
#c=sqrt(1+1)#
#c=sqrt(2)#
Compute #s#
#s=1/2*(a+b+c)#
#s=1/2*(sqrt(65)+sqrt(61)+sqrt(2))#
Compute r
#r = sqrt(((1/2*(sqrt(65)+sqrt(61)+sqrt(2))-sqrt(65))(1/2*(sqrt(65)+sqrt(61)+sqrt(2))-sqrt(61))(1/2*(sqrt(65)+sqrt(61)+sqrt(2))-sqrt(2)))/(1/2*(sqrt(65)+sqrt(61)+sqrt(2))))" "#
#r=sqrt(((0.581102745)(0.8331108174)(7.229146931))/8.643360493)#
#r=0.6363265774#
God bless.....I hope the explanation is useful.
