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Given the equation,
\(x ^2 +(x−2y−1) ^2 =−4y(x+y) \)
\(⇒ x^ 2 +4xy+4y^ 2 +(x−2y−1)^ 2 =0\)
\(⇒ (x+2y) ^2 +(x−2y−1) ^2 =0\)
As squares of real numbers are non-negative, the sum of two squares equals zero if and only if each square term is zero.
\(x - 2y - 1 = 0 \)
\(⇒ x - 2y = 1\)