AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

The point of intersection of the lines x+y+1=0 and
2x−3y+4=0 is A(−

7

5

,

2

5

)
If the coordinate of the image of point P(−2,0) with respect to
line x+y+1=0 is (h,k), then

h+2

1

=

k−0

1

=2×

−2+1

2

=1
So
h=−1
k=1
Thus (h,k)=(−1,1)
The equation of required bisector of angle is line joining the
points‌A(

−7

5

,

2

5

) and B(−1,1) is
y−1=

3 5

2 2

(x+1)
2y−2=3x+3
3x−2y+5=0