Q. 325.0( 1 Vote )

# A point equidista

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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