Given equations are AB 4x + 3y + 10 = 0
Normalizing AB, we get
= > 4x + 3y + 10 = 0
Dividing by 5, we get
\(\frac{4x}{5}+\frac{3y}{5}+2\) = 0 ……(1)
Consider BC 5x - 12y + 26 = 0
Normalizing BC we get,
= > \(\frac{15x}{13}-\frac{12y}{13}+2\) = 0 ……(2)
Consider AC 7x + 24y - 50 = 0
Normalizing AC we get
= > \(\frac{7x}{25}+\frac{24y}{25}-2\) = 0 ……(3)
Adding (1) + (3), we get Angular bisector of A: \(\frac{27x}{25}+\frac{39y}{25}\) = 0 ……(4)
Adding (2) + (3), we get Angular bisector of C: = 0 \(\frac{216x}{325}+\frac{12y}{325}\) = 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.
