2004 AMC 10A Problem 4

Since $|x - 1|$ and $|x - 2|$ are the distances from $x$ to $1$ and $2,$ the point $x$ is equidistant from $1$ and $2.$ That midpoint is $$x = \dfrac{1 + 2}{2} = \dfrac{3}{2}.$$ Thus, the correct answer is D.