# Inscribed circle

Write the equation of a incircle of the triangle KLM if K [2,1], L [6,4], M [6,1].

Result

x =  5
y =  2
r =  1

#### Solution:

$k=\sqrt{ (6-6)^2+(4-1)^2 }=3 \ \\ l=\sqrt{ (2-6)^2+(1-1)^2 }=4 \ \\ m=\sqrt{ (2-6)^2+(1-4)^2 }=5 \ \\ p=k+l+m=3+4+5=12 \ \\ s=p/2=12/2=6 \ \\ S=k \cdot \ l/2=3 \cdot \ 4/2=6 \ \\ \ \\ \ \\ x=(k \cdot \ 2+l \cdot \ 6+m \cdot \ 6)/p=(3 \cdot \ 2+4 \cdot \ 6+5 \cdot \ 6)/12=5$
$y=(k \cdot \ 1+l \cdot \ 4+m \cdot \ 1)/p=(3 \cdot \ 1+4 \cdot \ 4+5 \cdot \ 1)/12=2$
$r=S/s=6/6=1 \ \\ (x-5)^2+(y-2)^2=1$

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Math student
the answers are not proper. steps are missing.

Dr Math
you are right. The coordinates of the incenter are the weighted average of the coordinates of the vertices, where the weights are the lengths of the corresponding sides.

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