We call the corners vertices.
Let #r# be the radius of the incircle with incenter I. The perpendicular from I to each side is the radius #r#. That forms the altitude of a triangle whose base is a side. The three triangles together make the original trangle, so its area #mathcal{A}# is
# mathcal{A} = 1/2 r(a+b+c) #
We have
#a^2 = (9-5)^2 + (4-5)^2=17#
#b^2 = (9-1)^2 + (8-4)^2=80 #
# c^2 = (5-1)^2+(8-5)^2=25 #
The area #mathcal{A}# of a triangle with sides #a,b,c# satisfies
#16mathcal{A}^2 = 4a^2 b^2 - (c^2 - a^2 - b^2)^2 #
#16 mathcal{A}^2 = 4(17)(80) - (25 - 17 - 80)^2 = 256 #
#mathcal{A} = sqrt{256/16} = 4#
#r = {2 mathcal{A} } / (a+b+c) #
#r = { 8 } / {\sqrt{17} + sqrt{80} + sqrt{25} } #
