SSC CGL 16 Aug 2015 Shift 1 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
= $\frac{1}{2}\times 3\times 3-\frac{1}{2}$ × 3 × 2
= $\frac{9}{2}-\frac{6}{2}$ = $\frac{3}{2}$
= 1 $\frac{1}{2}$ sq. unit