For $x$ and $y$ satisfying $|x|+|y|=|x-3|+|y-2|$, which of the following is correct?

For $x$ and $y$ satisfying $|x|+|y|=|x-3|+|y-2|$, which of the following is correct?
  1. $x=\frac{1}{2}$ for $0 \leq x < 3$ and $1 \leq y < 2$
  2. $x+y=\frac{5}{2}$ for $x \geq 3$ and $y \geq 2$
  3. $x=\frac{1}{2}$ for $x \geq 2$ and $0 \leq y < 3$
  4. $x+y=\frac{5}{2}$ for $0 \leq x < 3,0 \leq y < 2$

Solution

$\because|x|+|y|=|x-3|+|y-2|$ Case I $0 \leq x < 3$ and $0 \leq y < 2$ $|x|+|y|=|x-3|+|y-2|$ $\Rightarrow \quad x+y=-(x-3)-(y-2)$ $\begin{aligned} & \Rightarrow \quad x+y=-x+3-y+2 \\ & \Rightarrow \quad 2 x+2 y=5 \Rightarrow x+y=5 / 2\end{aligned}$ Case II $x \geq 3$ and $y \geq 2$ $|x|+|y|=|x-3|+|y-2|$ $\Rightarrow \quad x+y=x-3+y-2$ $\Rightarrow \quad x=-5$ which is wrong. $\therefore$ Option (d) is true.

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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