TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the triangle shown below.The equation of line AB is calculated as,
y+1=

7+1

1+5

(x+5)
y+1=

4

3

(x+5)
4x−3y+17=0
The equation of line BC is calculated as,
y+1=

2+1

−1+5

(x+5)
y+1=

3

4

(x+5)
3x−4y+11=0
The equation of bisector of angle ABC is calculated as,

4x−3y+17

5

=+

3x−4y+11

5

4x−3y+17=3x−4y+11
x+y−6=0
Or

4x−3y+17

5

=−(

3x−4y+11

5

)
4x−3y+17=−3x+4y−11
7x−7y+28=0
x−y+4=0