SSC CGL 9 Aug 2015 Shift 1 Paper

Putting x = 0 in 3x + 4y = 12
3 × 0 + 4y = 12
⇒ y = $\frac{12}{4}$ = 3
∴ Point of intersection on y – axis = (0, 3)
Again, putting y = 0 in 3x + 4y = 12
3x + 4 × 0 = 12
Þ 3x = 12 Þ x = 4
∴ Point of intersection on x – axis = (4,0)
∴ OA = 4 and OB = 3
∴ AB = $\sqrt{\mathit{O}{\mathit{A}}^2+\mathit{O}{\mathit{B}}^2}$
= $\sqrt{4^3+3^2}$ = $\sqrt{16+9}$
= $\sqrt{25}$ = 5 units