SSC CGL Tier-1 16-08-2015 Shift 1 Solved Paper

x = 0 ⇒ Equation of y – axis
Putting x = 0 in 2x + 3y = 6
0 + 3y = 6 ⇒ y = 2
∴ Co-ordinates of point of intersection on y – axis
= (0, 2)
Again, putting y = 0, x = 3
∴ Point of intersection on x – axis = (3, 0)
In x + y = 3
Putting x = 0, y = 3
and on putting y = 0, x = 3
∴ Required area
= Δ OAC – Δ OAB
=

1

2

×3×3−

1

2

× 3 × 2
=

9

2

−

6

2

=

3

2

= 1

1

2

sq. unit