A triangle has corners at #(2 , 8 )#, #(5 ,7 )#, and #(3 ,1 )#. What is the radius of the triangle's inscribed circle?

1 Answer
Sep 30, 2017

midpoint of #(2,8)# and #(5,7)# = #(3.5,7.5)#
slope of line between #(2,8)# and #(5,7)# = #(8-7)/(2-5) = -1/3#
#:.# slope of perpendicular line # = - 1/(-1/3) = 3#
point slope equation:
#y-7.5 = 3(x-3.5)#
#y = 3x-10.5+7.5#
#y = 3x-3#

midpoint of #(2,8)# and #(3,1)# = #(2.5,4.5)#
slope of line between #(2,8)# and #(3,1)# = #(8-1)/(2-3) = -7#
#:.# slope of perpendicular line # = - 1/(-7) = 1/7#
point slope equation:
#y-4.5= 1/7(x-2.5)#
#y = 1/7x-5/14+4.5#
#y = 1/7x-34/7#

Now we get the system of equations:
#y = 3x-3#
#y = 1/7x-34/7#

Solving, we get #(-13/20,-99/20)#

The distance between #(-13/20,-99/20)# and #(3,1)# gives the radius of the circle.

#=sqrt(19490)/40 ~~ 6.98#