Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$ I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases, 
$C-1, \, x \in [2, \infty) $ 
So, $ x-1 + x-2 = 1 \Rightarrow x= 2$
$C-2, \, x \in [1, 2) $
$x-1 - x + 2 = 1 \Rightarrow 1 =1 \Rightarrow x\in [1,2) $
$C-3, \, x \in (- \infty, 1)$
$ - x + 1 - x+2 = 1 \Rightarrow x= 1 \notin (-\infty, 1) \Rightarrow x = \phi$ (null set)
Taking common of all three solution set, I get $x= \phi$ because of the last case. But the answer is supposed to be $x \in [1,2]$
But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$

Someone explain this too?

Alternative Solution
By the Triangle Inequality,
$$|x-1|+|x-2|=|x-1|+|2-x|\geq \big|(x-1)+(2-x)\big|=1\,.$$
The inequality becomes an equality if and only if $2-x=0$, or $x-1=\lambda(2-x)$ for some $\lambda\geq0$ (which gives, by the way, $x=\frac{2\lambda+1}{\lambda+1}\in[1,2)$).  It follows immediately that $[1,2]$ is the solution set for $x\in\mathbb{C}$ such that $|x-1|+|x-2|=1$.
