Finding the Area Bounded by the Absolute Value Inequality \( |x| + |y| \leq 2 \)

The problem asks for the area of the region defined by the inequality \( |x| + |y| \leq 2 \). This type of inequality involves absolute values, which represent the distance from zero. To understand the region, we can consider the boundary curve \( |x| + |y| = 2 \).

Understanding \( |x| + |y| = k \)

The equation \( |x| + |y| = k \) (where \( k > 0 \)) describes a square centered at the origin. We can analyze this equation by considering the four quadrants:

• Quadrant I (\( x \geq 0, y \geq 0 \)): The equation becomes \( x + y = 2 \). This is a line segment connecting (2,0) and (0,2).
• Quadrant II (\( x \leq 0, y \geq 0 \)): The equation becomes \( -x + y = 2 \). This is a line segment connecting (-2,0) and (0,2).
• Quadrant III (\( x \leq 0, y \leq 0 \)): The equation becomes \( -x - y = 2 \), or \( x + y = -2 \). This is a line segment connecting (-2,0) and (0,-2).
• Quadrant IV (\( x \geq 0, y \leq 0 \)): The equation becomes \( x - y = 2 \). This is a line segment connecting (2,0) and (0,-2).

These four line segments form the boundary of the region. The vertices of this shape are (2,0), (0,2), (-2,0), and (0,-2).

Identifying the Geometric Shape

The boundary \( |x| + |y| = 2 \) forms a square whose vertices lie on the x and y axes. The inequality \( |x| + |y| \leq 2 \) represents all points inside or on this square.

Calculating the Area of the Region

The region is a square with vertices (2,0), (0,2), (-2,0), and (0,-2). We can calculate the area in a few ways:

Method 1: Using Diagonals

The diagonals of the square lie along the x and y axes. The length of the diagonal along the x-axis is the distance between (-2,0) and (2,0), which is \( 2 - (-2) = 4 \). The length of the diagonal along the y-axis is the distance between (0,-2) and (0,2), which is \( 2 - (-2) = 4 \). Since the diagonals are perpendicular and equal, the shape is a square. The area of a square with diagonal length \(d\) is given by \( \frac{1}{2} d^2 \).

Area \( = \frac{1}{2} \times (4)^2 = \frac{1}{2} \times 16 = 8 \).

Method 2: Dividing into Triangles

The square can be divided into four congruent right-angled triangles, one in each quadrant, with vertices at the origin (0,0) and two adjacent vertices of the square on the axes. For example, the triangle in Quadrant I has vertices (0,0), (2,0), and (0,2). This is a right-angled triangle with base 2 and height 2.

Area of one triangle \( = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \).

Since there are four such triangles, the total area is \( 4 \times 2 = 8 \).

Method 3: Finding Side Length

The side length of the square can be found using the distance formula between two adjacent vertices, for instance, (2,0) and (0,2).

Side length \( s = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

The area of the square is \( s^2 \).

Area \( = (2\sqrt{2})^2 = (2)^2 \times (\sqrt{2})^2 = 4 \times 2 = 8 \).

All methods yield the same result for the area of the region bounded by \( |x| + |y| \leq 2 \).

Therefore, the area of the region bounded by \( |x| + |y| \leq 2 \) is 8 square units.

Revision Table: Absolute Value Area Calculation

Additional Information: Graphing Absolute Value Inequalities

Graphing inequalities involving absolute values like \( |x| + |y| \leq k \) requires understanding how the absolute value affects the coordinates in different quadrants. The equation \( |x| + |y| = k \) creates a shape that is symmetric with respect to the x-axis, the y-axis, and the origin. This is because replacing \(x\) with \( -x \) or \(y\) with \( -y \) or both does not change the equation.

The region \( |x| + |y| \leq k \) includes all points whose "Manhattan distance" (or taxicab distance) from the origin \( (0,0) \) is less than or equal to \( k \). The Manhattan distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( |x_1 - x_2| + |y_1 - y_2| \). So, \( |x| + |y| \leq 2 \) means all points \( (x,y) \) such that their Manhattan distance from \( (0,0) \) is at most 2.

For a general inequality \( |x| + |y| \leq k \), the area is \( \frac{1}{2} (2k)^2 = \frac{1}{2} (4k^2) = 2k^2 \). In this problem, \( k=2 \), so the area is \( 2 \times (2)^2 = 2 \times 4 = 8 \).