2004 AMC 10A Problem 4

1. The expression $|x - 1|$ is the distance on the number line between $x$ and $1$. In the same way, $|x - 2|$ is the distance between $x$ and $2$.
2. The equation says these two distances are the same. The only point that is equally far from $1$ and from $2$ is the point exactly halfway between them:
$$x = \frac{1+2}{2} = \frac32.$$
3. Check: $\left|\tfrac32 - 1\right| = \tfrac12$ and $\left|\tfrac32 - 2\right| = \tfrac12$. They match.

The answer is $\boxed{\textbf{(D)}\ \frac32}$.

The absolute value of a difference, $|a - b|$, is the distance between $a$ and $b$. So an equation like $|x - p| = |x - q|$ asks for the point equally far from $p$ and $q$, which is their midpoint. Thinking of distances is quicker and safer than splitting into cases.

Two numbers have the same absolute value when they are equal or when one is the negative of the other.

• Equal: $x - 1 = x - 2$ would mean $-1 = -2$, which is false.
• Opposite: $x - 1 = -(x - 2) = 2 - x$. Then $2x = 3$, so $x = \tfrac32$.

On the number line, $|x-1|$ measures how far $x$ is from $1$, and $|x-2|$ measures how far $x$ is from $2$. The equation says these distances are equal, so $x$ sits exactly halfway between $1$ and $2$:
$$x = \frac{1+2}{2} = \frac32.$$

Check: $\left|\tfrac32 - 1\right| = \tfrac12$ and $\left|\tfrac32 - 2\right| = \tfrac12$.

The answer is $\boxed{\textbf{(D)}\ \frac32}$.