5・11 The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1 .
(B) 2 .
(C) $\pi$.
(D) 4 .
(China High School Mathematics League, 1982)
［Solution 1］The given equation can be transformed into the following four equivalent forms
$$
\begin{array}{c}
\left\{\begin{array}{l}
x-1 \geqslant 0, \\
y-1 \geqslant 0, \\
x-1+y-1=1 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x-1 \leqslant 0, \\
y-1 \geqslant 0, \\
1-x+y-1=1 ;
\end{array}\right. \\
\left\{\begin{array} { l } 
{ x - 1 \leqslant 0 , } \\
{ y - 1 \leqslant 0 , } \\
{ 1 - x + 1 - y = 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x-1 \geqslant 0, \\
y-1 \leqslant 0, \\
x-1+1-y=1 .
\end{array}\right.\right.
\end{array}
$$

By plotting the graph, the area can be determined to be 2.
[Solution 2] Using coordinate translation
$$
\left\{\begin{array}{l}
x^{\prime}=x-1, \\
y^{\prime}=y-1,
\end{array}\right.
$$

then the original equation becomes $\left|x^{\prime}\right|+\left|y^{\prime}\right|=1$.
It is easy to see that the area enclosed by the figure is 2 (note that coordinate translation does not change the size of the area of the figure).
Therefore, the answer is $(B)$.