Answer :

Given equations are AB 4x + 3y + 10 = 0


Normalizing AB, we get


= > 4x + 3y + 10 = 0


Dividing by 5, we get


……(1)


Consider BC 5x - 12y + 26 = 0


Normalizing BC we get,


= > ……(2)


Consider AC 7x + 24y - 50 = 0


Normalizing AC we get


= > ……(3)


Adding (1) + (3), we get Angular bisector of A: ……(4)


Adding (2) + (3), we get Angular bisector of C: = 0 ……(5)


Finding point of intersection of lines (4) and (5), we get I(0, 0) which is the


Incenter of the given triangle which is the point equidistant from its sides of a triangle.

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