A triangle has corners at #(4 , 5 )#, #(1 ,3 )#, and #(3 ,4 )#. What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-01-30T13:07:28

Radius of Incircle #color(red)(r = 0.2224)#

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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#