Q. 325.0( 1 Vote )
A point equidista A. (1, -1)
B. (1, 1)
C. (0, 0)
D. (0, 1)
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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