Question:
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?
Show Hint
For equations involving absolute values, divide the plane into regions according to the sign changes of the expressions inside the modulus and simplify separately in each region.
Updated On: Jun 18, 2026
- \[ x=\frac12 \] for \[ 0\le x<3,\quad 1\le y<2 \]
- \[ x+y=\frac52 \] for \[ x\ge 3,\quad y\ge 2 \]
- \[ x=\frac12 \] for \[ x\ge 2,\quad 0\le y<3 \]
- \[ x+y=\frac52 \] for \[ 0\le x<3,\quad 0\le y<2 \]
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Consider the region \(0\le x<3\) and \(0\le y<2\).
In this region, \[ |x|=x, \] \[ |y|=y, \] \[ |x-3|=3-x, \] and \[ |y-2|=2-y. \] Substituting into the given equation, \[ x+y=(3-x)+(2-y). \]
Step 2: Simplify the equation.
\[ x+y=5-x-y. \] \[ 2x+2y=5. \] \[ x+y=\frac52. \] Thus, throughout the region \[ 0\le x<3,\quad 0\le y<2, \] the given condition reduces to \[ x+y=\frac52. \]
Step 3: Verify the other options.
Option (1) claims \[ x=\frac12. \] However, the equation obtained is \[ x+y=\frac52, \] not a fixed value of \(x\). Hence option (1) is false.
Option (2) considers \[ x\ge3,\quad y\ge2. \] Then \[ |x|=x,\quad |y|=y, \] \[ |x-3|=x-3,\quad |y-2|=y-2. \] The equation becomes \[ x+y=x+y-5, \] which is impossible. Hence option (2) is false.
Option (3) also gives a fixed value \(x=\frac12\), which does not follow from the equation. Hence it is false.
Step 4: Final conclusion.
Therefore, the correct statement is \[ \boxed{x+y=\frac52} \] for \[ 0\le x<3,\quad 0\le y<2. \]
In this region, \[ |x|=x, \] \[ |y|=y, \] \[ |x-3|=3-x, \] and \[ |y-2|=2-y. \] Substituting into the given equation, \[ x+y=(3-x)+(2-y). \]
Step 2: Simplify the equation.
\[ x+y=5-x-y. \] \[ 2x+2y=5. \] \[ x+y=\frac52. \] Thus, throughout the region \[ 0\le x<3,\quad 0\le y<2, \] the given condition reduces to \[ x+y=\frac52. \]
Step 3: Verify the other options.
Option (1) claims \[ x=\frac12. \] However, the equation obtained is \[ x+y=\frac52, \] not a fixed value of \(x\). Hence option (1) is false.
Option (2) considers \[ x\ge3,\quad y\ge2. \] Then \[ |x|=x,\quad |y|=y, \] \[ |x-3|=x-3,\quad |y-2|=y-2. \] The equation becomes \[ x+y=x+y-5, \] which is impossible. Hence option (2) is false.
Option (3) also gives a fixed value \(x=\frac12\), which does not follow from the equation. Hence it is false.
Step 4: Final conclusion.
Therefore, the correct statement is \[ \boxed{x+y=\frac52} \] for \[ 0\le x<3,\quad 0\le y<2. \]
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