The point $(1 , 2)$ and the line $3x - 4y - 5 = 0$ are on the $xy$ -plane.

What is the shortest distance between the line and the point?

The answer is 2.

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Given the point $(m , n)$ and the line $Ax + By + C = 0$ , the distance between the point and the line can be formulated as: $\dfrac{|Am + Bn + C|}{\sqrt{A^2 + B^2}}$

As a result, the distance between the point $(1 , 2)$ and the line $3x - 4y - 5 = 0$ equals to $\dfrac{|3\times 1 - 4\times 2 - 5|}{\sqrt{3^2 + 4^2}}$ = $\dfrac{|-10|}{5} = \boxed{2}$