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2004 AMC 10A Problems/Problem 4

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Problem

What is the value of $x$ if $|x-1|=|x-2|$?

$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 {2}\qquad \mathrm{(E) \ } 2$

Solution

$|x-1|$ is the same thing as saying the distance between $x$ and $1$; $|x-2|$ is the same thing as saying the distance between $x$ and $2$.

Therefore, $x$ is the same distance from $1$ and $2$, so $x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}$.

See Also

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