For x and y satisfying |x|+|y|=|x-3|+|y-2|,wh | vedclass

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(D) Given equation: $|x|+|y|=|x-3|+|y-2|$.Case $I$: $0 \leq x < 3$ and $0 \leq y < 2$.In this region,$|x|=x$,$|y|=y$,$|x-3|=-(x-3)$,and $|y-2|=-(y-2)$

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$x+y=\frac{5}{2}$ for $0 \leq x < 3, 0 \leq y < 2$

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(D) Given equation: $|x|+|y|=|x-3|+|y-2|$.Case $I$: $0 \leq x In this region,$|x|=x$,$|y|=y$,$|x-3|=-(x-3)$,and $|y-2|=-(y-2)$.Substituting these into the equation: $x+y = -(x-3) - (y-2)$.$x+y = -x+3-y+2$.$2x+2y = 5$.$x+y = 5/2$.Case $II$: $x \geq 3$ and $y \geq 2$.In this region,$|x|=x$,$|y|=y$,$|x-3|=x-3$,and $|y-2|=y-2$.Substituting these into the equation: $x+y = (x-3) + (y-2)$.$x+y = x+y-5$.$0 = -5$,which is impossible.Therefore,the condition $x+y = 5/2$ holds for $0 \leq x < 3$ and $0 \leq y < 2$.

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

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