Solution:
Let $x = A\sin(\omega t)$ and $y = B\sin(\omega t + \pi) = -B\sin(\omega t)$.\nThen $y/x = -B/A \implies y = -(B/A)x$.\nThis represents the equation of a straight line.
The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along: (1990)
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