Question

Find the incentre of the triangle formed by the vertices (0,0). (5.-1) and (-2,3).​

Answer 1

incentre of the triangle is = [(5 - 2√2)/(√5 + 1 + √2), (3√2 - 1)/(√5 + 1 + √2)]

first find side lengths of triangle.

if A = (0,0) , B = (5, -1) , C = (-2,3)

then, a = BC = √{(5 + 2)² + (-1 - 3)²} = √{49 + 16} = √65

b = CA = √{(-2-0)² + (3 - 0)²} = √13

c = AB = √{(5 - 0)² + (-1-0)²} = √26

now incentre = [(ax1 + bx2 + cx3)/(a + b + c), (ay1 + by2 + cy3)/(a + b + c)]

= [(√65 × 0 + √13 × 5 + √26 × -2)/(√65 + √13 + √26) , (√65 × 0 + √13 × -1 + √26 × 3 )/(√65+ √13 + √26)]

= [(5√13 - 2√26)/(√65 + √13 + √26), (3√26 - √13)/(√65 + √13 + √26)]

= [(5 - 2√2)/(√5 + 1 + √2), (3√2 - 1)/(√5 + 1 + √2)] [ you can rationalize it, you will get another form of it. ]

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Answer 2

by the above process we can do it

I think it helps you