Straight Lines – Intersection of lines – JEE Main 23 January 2025 Shift 2

Solution:

Given sets:
$$A = \{(x,y) : |x+y| \ge 3\}$$
$$B = \{(x,y) : |x| + |y| \le 3\}$$
$$C = \{(x,y) \in A \cap B : x=0 \text{ or } y=0\}$$

We need to find points in $C$. The condition $x=0$ or $y=0$ means the points lie on the coordinate axes.

**Case 1:** Points on the Y-axis ($x=0$).
Substitute $x=0$ into the inequalities for sets $A$ and $B$:
From set $B$: $|0| + |y| \le 3 \Rightarrow |y| \le 3 \Rightarrow -3 \le y \le 3$.
From set $A$: $|0+y| \ge 3 \Rightarrow |y| \ge 3$.

Combining these conditions ($|y| \le 3$ and $|y| \ge 3$), we get $|y| = 3$.
So, $y = 3$ or $y = -3$.
Points obtained: $(0, 3)$ and $(0, -3)$.

**Case 2:** Points on the X-axis ($y=0$).
Substitute $y=0$ into the inequalities for sets $A$ and $B$:
From set $B$: $|x| + |0| \le 3 \Rightarrow |x| \le 3 \Rightarrow -3 \le x \le 3$.
From set $A$: $|x+0| \ge 3 \Rightarrow |x| \ge 3$.

Combining these conditions ($|x| \le 3$ and $|x| \ge 3$), we get $|x| = 3$.
So, $x = 3$ or $x = -3$.
Points obtained: $(3, 0)$ and $(-3, 0)$.



The set $C$ contains the points:
$$C = \{(0, 3), (0, -3), (3, 0), (-3, 0)\}$$

We need to calculate $\sum_{(x,y)\in C}|x+y|$:
1. For $(0, 3)$: $|0 + 3| = 3$
2. For $(0, -3)$: $|0 – 3| = 3$
3. For $(3, 0)$: $|3 + 0| = 3$
4. For $(-3, 0)$: $|-3 + 0| = 3$

Total Sum $= 3 + 3 + 3 + 3 = 12$.

Ans. (12)