A point equidistant from the line 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is
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.