Question

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

Answer 1

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

Explanation:

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`