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We have, \frac{(1 + i) x - i}{2 + i} + \frac{(1 + 2 i) y - i}{2 - i} = 1 \Rightarrow \frac{(1 + i) (2 - i) x - i (2 - i) + (1 + 2 i) (2 + i) y - i (2 + i)}{(2 + i) (2 - i)} = 1 \Rightarrow \frac{2 x + ix + x - 2 i - 1 + 2 y + 5 iy - 2 y - 2 i + 1}{4 + 1} = 1 \Rightarrow 3 x + i (x - 4 + 5 y) = 5 On comparing real and imaginary parts, we get 3 x = 5 and x - 4 + 5 y = 0 x = \frac{5}{3} and \frac{5}{3} - 4 + 5 y = 0 \Rightarrow 5 y = 4 - \frac{5}{3} \Rightarrow 5 y = \frac{12 - 5}{3} \Rightarrow y = \frac{7}{15} So, (x, y) = (\frac{5}{3}, \frac{7}{15}) There is a printing mistake in the options given .

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