SSC CHSL 2024 PYQ — Question 63

The correct answer is option 2 i.e. 4.8 unit.

Perpendicular distance from a point (x₁, y₁) to a line Ax + By + C = 0 $=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

Here, x₁ = 0, y₁ = 0 and A = 6, B = 8, C = -48

Distance $=\frac{|6\times 0+8\times 0-48|}{\sqrt{6^2+8^2}}$

$=\frac{|-48|}{\sqrt{36+64}}$

$=\frac{48}{\sqrt{100}}$

= 48/10

= 4.8