For a triangle with Cartesian vertices# (x_1,y_1), (x_2,y_2), (x_3,y_3),# the Cartesian coordinates of the incenter are given by
#(x_I,y_I)=((ax_1+bx_2+cx_3)/(a+b+c),(ay_1+by_2+cy_3)/(a+b+c)).#
First lets find the lengths of a, b, c.
#a = sqrt((4-1)^2 + (5-3)^2) = 3.6056#
#b = sqrt((1-3)^2 + (3-4)^2) = 2.2361#
#c = sqrt((3-4)^2 + (4-5)^2) = 1.4142#
#a + b + c = 7.2559#
#x_I = ((3.6056*4) + (2.2361*1) + (1.4142*3)) / 7.2559 = 2.8806#
#y_I = ((3.6056*5)+(3*2.2361)+(4*1.4142))/7.2559 = 4.1887#
Slope side a #m_a = (y_3 - y_2) / (x_3 - x_2) = (4-3) / (3-1) = 1/2#
Slope of IM_A perpendicular to side 'a' #m_(I-M_A) = -1 / (1/2) = -2#
Equation of #IM_A# is
#y - 4.1887 = -2 (x - 2.8806)#
#y + 2x = 4.1887 + 5.7612 = 9.9499# Eqn (1)
Equation of BC = side 'a' is
(y - 3) / (4-3) = (x - 1) / (3-1)#
#2y -x = 5# Eqn (2)
Solving Eqns (1), (2) we get the coordinates of #M_A#
#M_A (3,4)#
Radius of Incircle #r = IM_A = IM_B = IM_C#
#I_ = sqrt((3 - 2.8806)^2 + (4-4.1887)^2) ~~ 0.2224#