Given sides
x−2y+4=0-----(1) and
2x+y−5=0-----(2)
from eq (1) and (2) we get intersecting point
I(65,135)
Eq of angle bisector of given lines
x−2y+4=±(2x+y−5)
on solving we get
x+3y=9 and 3x−y=1
Side BC will be parallel to these bisectors.Let
AD=a where D is foot of perpendicular
⇒AB=a√2 and area of ΔABC is
12×(a√2)2=a2=10
⇒a=√10
Let eq of BC be x+3y=λ
√10=65−395−λ√10⇒λ=−1,19
Eq of BC is x+3y=−1 or x+3y=19
If eq of BC is 3x−y=λ
√10=185−135−λ√10⇒λ=−9,11
Eq of BC is 3x−y=−9 or 3x−y=11