Question

find the incentre of Triangle formed by the vertices (0,0)(5,-1),(-2,3)​

Answer 1

(5 - 2√2)/(√5 + 1 + √2)   ,  ( 3√2 - 1)/(√5 + 1 + √2) are incenter of  Triangle formed by the vertices (0,0)(5,-1),(-2,3)​

Step-by-step explanation:

Let say

A = (0 , 0)  

B = ( 5 , -1)

C = (- 2 , 3)

a  = BC  =  √7² + 4² = √65

b = AC = √13

c = AB =  √26

a + b + c =  √65 + √13 + √26  = √13(√5 + 1 + √2)

incenter  =  (  √65(0) +  √13*5  +  √26(-2)/√13(√5 + 1 + √2)) ,

( (  √65(0) +  √13*(-1)  +  √26(3)/√13(√5 + 1 + √2))

= (5 - 2√2)/(√5 + 1 + √2)   ,  ( 3√2 - 1)/(√5 + 1 + √2)

Learn more:

In a triangle abc the incircle touches the sides bc,ca and ab ...

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