If the sides of a triangle are #a#, #b# and #c#, then the area of the triangle #Delta# is given by the formula
#Delta=sqrt(s(s-a)(s-b)(s-c))#, where #s=1/2(a+b+c)#
and radius of inscribed circle is #Delta/s#
Hence let us find the sides of triangle formed by #(5,2)#, #(9,9)# and #(6,8)#. This will be surely distance between pair of points, which is
#a=sqrt((9-5)^2+(9-2)^2)=sqrt(16+49)=sqrt65=8.0623#
#b=sqrt((6-9)^2+(8-9)^2)=sqrt(9+1)=sqrt10=3.1623# and
#c=sqrt((6-5)^2+(8-2)^2)=sqrt(1+36)=sqrt37=6.0828#
Hence #s=1/2(8.0623+3.1623+6.0828)=1/2xx17.3074=8.6537#
and #Delta=sqrt(8.6537xx(8.6537-8.0623)xx(8.6537-3.1623)xx(8.6537-6.0828)#
= #sqrt(8.6537xx0.5914xx5.3914xx2.5709)=sqrt70.9365=8.4224#
And radius of inscribed circle is #8.4224/8.6537=0.9733#
