NIMCET 2022 — Mathematics PYQ

 The equation $|x|+|y|=2$ represents a closed figure.
   We split it into four cases depending on the signs of $x$ and $y$:

* Quadrant I ($x\ge 0,y\ge 0$): $x+y=2$
* Quadrant II ($x\le 0,y\ge 0$): $-x+y=2\Rightarrow y=2+x$
* Quadrant III ($x\le 0,y\le 0$): $-x-y=2\Rightarrow x+y=-2$
* Quadrant IV ($x\ge 0,y\le 0$): $x-y=2\Rightarrow y=x-2$

2. These four lines intersect at points $(2,0),(0,2),(-2,0),(0,-2)$.
   So the curve forms a **diamond (a square rotated 45°)**.

3. To find the area, we can use two simple methods:

   **Method A (Triangle method):**
   In Quadrant I, the line $x+y=2$ with the axes forms a right triangle with vertices $(0,0),(2,0),(0,2)$.
   Its area = $\frac{1}{2}\times 2\times 2=2$.
   There are 4 such identical triangles → total area = $4\times 2=8$.

   **Method B (Diagonal method):**
   The diamond has diagonals joining $(2,0)$ to $(-2,0)$ and $(0,2)$ to $(0,-2)$.
   Both diagonals = $4$.
   Area of a rhombus = $\frac{d_1{d}_2}{2}=\frac{4\cdot 4}{2}=8$.

4. Therefore, the required area = **8 sq. units**