6. In the Cartesian coordinate system, the equation
$\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1 \quad(a, b$ are two unequal positive numbers $)$ represents the curve of
(A) Triangle;
(B) Square;
(C) Rectangle that is not a square;
(D) Rhombus that is not a square.
6. (D)

Rotate the Cartesian coordinate system $X O Y$ counterclockwise by $45^{\circ}$ to obtain the new coordinate system $X^{\prime} O Y^{\prime}$. Point $P$ has coordinates $\left(x^{\prime}, y^{\prime}\right)$ in the coordinate system $X^{\prime} O Y^{\prime}$ and coordinates $(x, y)$ in the coordinate system $X O Y$. Then,
$$
\left\{\begin{array}{l}
x^{\prime}=\frac{1}{\sqrt{2}}(x+y), \\
y^{\prime}=\frac{1}{\sqrt{2}}(-x+y) .
\end{array}\right.
$$

The equation in the problem
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$

transforms into
$$
\frac{\left|x^{\prime}\right|}{a}+\frac{\left|y^{\prime}\right|}{b}=\sqrt{2}
$$

Clearly, (2) represents a curve that is symmetric with respect to the $X^{\prime}$-axis and the $Y^{\prime}$-axis. In the first quadrant of $X^{\prime} O Y^{\prime}$, (2) becomes $\frac{x^{\prime}}{a}+\frac{y^{\prime}}{b}=\sqrt{2}$,

which is the line segment $A B$, where $A(\sqrt{2} a, 0), B(0, \sqrt{2} b)$.

By symmetry, in the second quadrant, equation (2) is the line segment $B C$, where $C(-\sqrt{2} a, 0)$;
in the third quadrant, equation (2) is the line segment $C D$, where $D(0,-\sqrt{2} b)$;
in the fourth quadrant, equation (2) is the line segment $A D$.
By symmetry, $A B=B C=C D=A D$, and since $a \neq b$, $A C \neq B D$, so $A B C D$ is a rhombus that is not a square.