SSC CGL 19 Oct 2014 Shift 2 Paper

x = 4, a straight line parallel to y – axis.
y = 3, a straight line parallel to x – axis.
Putting x = 0 in 3x + 4y = 12,
3 × 0 + 4y = 12,
⇒ 4y = 12 ⇒ y = $\frac{12}{3}$ = 4
∴ Point of intersection on y – axis = (0, 4)
Again, putting y = 0 in 3x + 4y = 12,
3x + 4 × 0 = 12
⇒ 3x = 12 ⇒ x = $\frac{12}{3}$ = 4
∴ Point of intersection on x - axis = (4, 0)

Area of  ΔACB = OA × OB
= 4 × 3 = 12 sq. units
Area of Δ OAB = $\frac{1}{2}$ × OA × OB = $\frac{1}{2}$ × 4 × 3 = 6 sq. units
∴ Area of Δ ABC = 12 – 6 = 6 sq. units