JAMB 1992 · UME · Q32

What is the perpendicular distance of the point $(2, 3)$ from the line $2x - 4y + 3 = 0$ ?

Worked solution (try it first)

The distance from $(x_1, y_1)$ to $ax + by + c = 0$ is $\dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ .
Put in $(2, 3)$ : the top is $|2(2) - 4(3) + 3| = |4 - 12 + 3| = 5$ .
The bottom is $\sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt5$ .
So the distance is $\frac{5}{2\sqrt5} = \frac{\sqrt5}{2}$ , option A.