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A square of side ‘a’ lies above the x-axis and has one vertex at the origin. The side passing through
the origin makes an angle $α (0 < α < π / d)$ with the positive direction of x-axis. The equation of
its diagonal not passing through the origin is

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$y (\cos α + \sin α) + x (\sin α - \cos α) = a$

$y (\cos α + \sin α) - x (\sin α - \cos α) = a$

$y (\cos α - \sin α) - x (\sin α - \cos α) = a$

$y (\cos α + \sin α) + x (\sin α + \cos α) = a$

answer is B.

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Detailed Solution

Slope of BC $\tan (\frac{π}{4} + α)$

Slope of CA $- \cot (\frac{π}{4} + α)$

Equation of the $CA (y - a \sin α) = - \cot (\frac{π}{4} + α) (x - a \cos α)$

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