Question

the in-centre of the triangle with with vertices ,(0,0),(1,0),(0,1) is​

Answer 1

A=(0,0)

B=(1,0)

C=(0,1)

AB=√(0-1)^2+(0-0)^2

→AB=1

→c=1

BC=√(1-0)^2+(0-1)^2

→BC=√2

→a=√2

CA=√(0-0)^2+(0-1)^2

→CA=1

→b=1

Now,

Incentre=

({ax1+bx2+cx3}/{a+b+c},{ay1+by2+cy3}/{a+b+c}

On putting required value and solving it we get

→(1/(2-√2),(1/(2-√2))

→({2-√2}/2,{2+√2}/2)

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