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) Non-square rectangle;
(D) Non-square rhombus.
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 $\left(x^{\prime}\right.$, $\left.y^{\prime}\right)$ 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 equations in the problem are:
$$
\begin{array}{c}
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1 \\
\frac{\left|x^{\prime}\right|}{a}+\frac{\left|y^{\prime}\right|}{b}=\sqrt{2}
\end{array}
$$

Clearly, (2) represents a curve that is symmetric about 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 $A(\sqrt{2} a, 0)$ and $B(0, \sqrt{2} b)$.

By symmetry, in the first quadrant, equation (2) is the line segment $B C$, where $C(-\sqrt{2} a, 0)$;
in the first 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.