Question

O(0,0),A(6,0)B(0,8) are vertices of a triangle find the coordinates of the incentre of triangle OAB​

Answer 1

Given :  vertices O(0,0),A(6,0)and B(0,8) of ∆OAB

To Find : The coordinates of the point equidistant from the vertices

Solution:

O(0,0),A(6,0)and B(0,8)  

(h , k)  coordinates of the point equidistant from the vertices

=> ( h - 0)² + (k - 0)² = (h - 6)² + ( k - 0)²  = ( h - 0)² + ( k - 8)²

( h - 0)² + (k - 0)² = (h - 6)² + ( k - 0)²

=> h²  = h² - 12h + 36

=> h = 3

( h - 0)² + (k - 0)² =   ( h - 0)² + ( k - 8)²

=> k² = k² -16k + 64

=> k = 4

(h , k)

( 3 , 4)

The coordinates of the point equidistant from the vertices O(0,0),A(6,0)and B(0,8) of ∆OAB are ( 3 , 4)

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