Denote the lengths of the sides of the triangle as:
#color(white)("XXX")a#: between #(3,3)# and #(1,2)#
#color(white)("XXX")b#: between #(1,2)# and #(5,9)#
#color(white)("XXX")c#: between #(5,9)# and #(3,3)#
and also
#color(white)("XXX")s# as the semi-perimeter #=(a+b+c)/2#
Then the radius of the triangle's inscribed circle is
#color(white)("XXX")r=("Area"_triangle)/s#
or using Heron's formula
#color(white)("XXX")r=(sqrt(s(s-a)(s-b)(s-c)))/s#
Using the Pythagorean Theorem (and a calculator)
#color(white)("XXX")a=sqrt((3-1)^2+(3-2)^2)~~2.236#
#color(white)("XXX")b=sqrt((1-5)^2+(2-9)^2)~~8.062#
#color(white)("XXX")c=sqrt((5-3)^2+(9-3)^2)~~6.325#
#color(white)("XXX")s~~8.311#
#"Area"_triangle=sqrt(s(s-a)(s-b)(s-c))=5#
#"radius of incircle" = 5/8.311 ~~0.602#
