The region represented by |x-y| ≤ 2 and |x+y| ≤ 2 is bounded by a : (A) square of side length 2√2 un

Q: Wait, why do we treat $|x-y| \le 2$ as two lines? Isn't it just one region?

Great question! Think of the modulus as a distance. $|x-y| \le 2$ means the distance between $x$ and $y$ is at most 2. On the coordinate plane, this defines a region trapped between the lines $x-y = 2$ and $x-y = -2$. It's not just one line; it's the entire 'corridor' between them. When you have an inequality, you are looking at the area *between* the boundaries, not just the boundaries themselves.

Q: How do I know if it's a square or a rhombus? Do I have to check the angles?

You are absolutely right to be cautious! A rhombus is a quadrilateral with all sides equal, but a square is a special type of rhombus where the angles are also $90^\circ$. Since we found the side lengths are equal, we know it's a rhombus. To confirm it's a square, we must check the slopes of the adjacent sides. If the product of the slopes is $-1$, the sides are perpendicular, confirming it's a square.

Q: Is there a faster way to visualize this without plotting all four lines?

Visualization is a skill you build with practice! You can think of the transformation $u = x-y$ and $v = x+y$. The inequalities become $|u| \le 2$ and $|v| \le 2$. This is a square in the $uv$-plane. Since the transformation is a rotation and scaling, it remains a square in the $xy$-plane. But for now, plotting the lines is the most reliable way to ensure you don't miss any details!

Question

The region represented by $|x-y| \le 2$ and $|x+y| \le 2$ is bounded by a :

Options:

(A) square of side length $2\sqrt{2}$ units

(B) rhombus of side length 2 units

(C) square of area 16 sq. units

(D) rhombus of area $8\sqrt{2}$ sq. units

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Constraints

Welcome, fellow traveler of the coordinate plane! Today, we are not just solving an inequality; we are embarking on a journey to visualize the hidden architecture of the Cartesian plane.

When you see expressions like $|x-y| \le 2$ and $|x+y| \le 2$, do not just see algebra. See the boundaries of a world waiting to be defined.

### The Modulus Corridor

Let us begin with the first inequality: $|x-y| \le 2$. In the language of mathematics, the modulus property $|X| \le a$ tells us that $-a \le X \le a$.

Applying this to our expression, we get:

$$-2 \le x-y \le 2$$

This is not just a single line; it is a vast, infinite corridor trapped between two parallel lines: $x-y = 2$ and $x-y = -2$. Imagine standing on the $xy$-plane and drawing these two lines. Everything between them is part of the solution set for this first constraint.

### The Intersection of Worlds

Now, we introduce the second constraint: $|x+y| \le 2$. By the same logic, this expands to $-2 \le x+y \le 2$, giving us another pair of parallel lines: $x+y = 2$ and $x+y = -2$.

Now, we have two corridors crossing each other. The region that satisfies **both** inequalities simultaneously is the intersection of these two strips. It is the heart of the overlap, a closed quadrilateral sitting right at the origin of our coordinate system.

### Unveiling the Vertices

To truly understand this shape, we must find its corners. We find the vertices by solving the systems of equations formed by the intersection of these lines.

For instance, the rightmost vertex is the intersection of $x-y = 2$ and $x+y = 2$. Adding these equations, we get:

$$2x = 4 \Rightarrow x = 2$$

Substituting back, we find $y = 0$. Our first vertex is $(2, 0)$. By repeating this process for the other intersections, we find the remaining vertices: $(0, 2)$, $(-2, 0)$, and $(0, -2)$. We have successfully mapped the four corners of our quadrilateral.

### The Proof of the Square

Now, we must identify the nature of this shape. We calculate the side length using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Taking the distance between $(2, 0)$ and $(0, 2)$, we get:

$$d = \sqrt{(0-2)^2 + (2-0)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$

Since all sides are equal, we have a rhombus. But is it a square? We check the slopes. The slope of $x-y=2$ is $m_1 = 1$, and the slope of $x+y=2$ is $m_2 = -1$.

Their product is:

$$m_1 	imes m_2 = 1 	imes (-1) = -1$$

This is the golden ticket! The sides are perpendicular, meaning the angles are $90^\circ$. A rhombus with $90^\circ$ angles is, by definition, a square.

### Conclusion

We have traversed the coordinate plane, defined our corridors, identified our vertices, and proven the geometric nature of our region. The result is a beautiful **square of side length $2\sqrt{2}$**.

Remember, in JEE Advanced, the math is the tool, but the visualization is the key. Keep exploring, keep questioning, and keep falling in love with the elegance of geometry!

Animated Solution for Mathematics - Straight Lines: The region represented by $∣ x - y ∣ \leq 2$ and $∣ x + y ∣ \leq 2$ is bounded by a :

Select Answer:

Visualized Solution

The Coordinate Plane

Expanding the First Inequality $∣ x - y ∣ \leq 2$

The Second Boundary Strip $∣ x + y ∣ \leq 2$

Identifying the Bounded Region

Finding the Vertices: Point $(2, 0)$

Finding the Vertices: Point $(0, 2)$

Finding the Remaining Vertices

Calculating the Side Length

Checking the Angles

Final Conclusion

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Geometry of Constraints

The Modulus Corridor

The Intersection of Worlds

Unveiling the Vertices

The Proof of the Square

Conclusion

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The region represented by ∣x−y∣≤2 and ∣x+y∣≤2 is bounded by a :