The area of the closed region bounded by the equation $| x | + | y | = 2$ in the two-dimensional plane is

When we see an equation like $|x| + |y| = 2$, the absolute value signs mean we need to consider different cases based on whether $x$ and $y$ are positive or negative. This equation represents four different linear equations depending on the signs of $x$ and $y$:

When $x \geq 0$ and $y \geq 0$: $x + y = 2$

When $x \geq 0$ and $y < 0$: $x + (-y) = 2$, so $x - y = 2$

When $x < 0$ and $y \geq 0$: $(-x) + y = 2$, so $-x + y = 2$

When $x < 0$ and $y < 0$: $(-x) + (-y) = 2$, so $-x - y = 2$

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To understand the shape, let's find where this curve intersects the axes:

X-intercepts (where $y = 0$):

$|x| + |0| = 2$

$|x| = 2$

So $x = 2$ or $x = -2$

Y-intercepts (where $x = 0$):

$|0| + |y| = 2$

$|y| = 2$

So $y = 2$ or $y = -2$

This gives us four key points: $(2, 0)$, $(-2, 0)$, $(0, 2)$, and $(0, -2)$.

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When we connect these four points, we get a square rotated 45° with vertices at:

$(2, 0)$ - rightmost point

$(0, 2)$ - topmost point

$(-2, 0)$ - leftmost point

$(0, -2)$ - bottommost point

This is a square because all four sides have equal length, and all angles are 90°.

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The fastest way to find the area is using the diagonal method:

Horizontal diagonal: from $(-2, 0)$ to $(2, 0) = 4$ units

Vertical diagonal: from $(0, -2)$ to $(0, 2) = 4$ units

Area formula for a rhombus/square:

Area = $\frac{1}{2} \times d_1 \times d_2$

Where $d_1$ and $d_2$ are the lengths of the diagonals.

Area = $\frac{1}{2} \times 4 \times 4 = 8$