∵|x|+|y|=|x−3|+|y−2| Case I :0≤x<3 and 0≤y<2 |x|+|y|=|x−3|+|y−2| ⇒x+y=−(x−3)−(y−2) ⇒x+y=−x+3−y+2 ⇒2x+2y=5 ⇒x+y=5∕2 Case II : x≥3 and y≥2 |x|+|y|=|x−3|+|y−2| ⇒x+y=x−3+y−2 ⇒x=−5 which is wrong. ∴ option (d) is true.