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

square of side length 2 \sqrt 2 units

rhombus of area 8 \sqrt 2 sq. units

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q30 with Solution

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Correct answer: B

$|x-y|\le 2 \quad \text{and} \quad |x+y|\le 2.$

These inequalities mean:

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

So the boundary lines are:

$x-y=2,\quad x-y=-2,\quad x+y=2,\quad x+y=-2.$

Intersection of $x-y=2$ and $x+y=2$ : Adding, $2x=4 \Rightarrow x=2$ Then, $2+y=2 \Rightarrow y=0$ So one vertex is $(2,0)$ .
Intersection of $x-y=2$ and $x+y=-2$ : Adding, $2x=0 \Rightarrow x=0$ Then, $0-y=2 \Rightarrow y=-2$ So another vertex is $(0,-2)$ .
Intersection of $x-y=-2$ and $x+y=2$ : Adding, $2x=0 \Rightarrow x=0$ Then, $0+y=2 \Rightarrow y=2$ So another vertex is $(0,2)$ .
Intersection of $x-y=-2$ and $x+y=-2$ : Adding, $2x=-4 \Rightarrow x=-2$ Then, $-2+y=-2 \Rightarrow y=0$ So the fourth vertex is $(-2,0)$ .

Thus the vertices are:

$(2,0),\ (0,2),\ (-2,0),\ (0,-2).$

These points form a square (diamond-shaped, i.e. a square tilted at $45^\circ$ ).

Let us compute the side length between $(2,0)$ and $(0,2)$ :

= \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}.$$ So the figure is a square of side length $2\sqrt{2}$. 4. Its area is $$\text{Area}=(2\sqrt{2})^2=8.$$ So: - Option A: rhombus of area $8\sqrt{2}$ — false - Option B: square of side length $2\sqrt{2}$ — true - Option C: square of area $16$ — false - Option D: rhombus of side length $2$ — false 5. Therefore, the correct option is: $$\boxed{\text{B}}$$