So
0
0
1
1 1
3
1
2
1
x
y
 = 0
This gives
(
)
1
2 y –3
x  = 0 or y = 3x,
which is the equation of required line AB Also, since the area of the triangle ABD is 3 sq units, we have
1
3
1
1 0
0
1
2
0
1
k
 = ± 3
This gives, 
3
3
2
−k
= ± , i e