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 $α (0 < α < \frac{π}{4})$ 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

: Let OA = OC = a

r = a and (0, 0) = (x₁ y₁)

$\begin{array}{l} A = (x_{1} + r \cos θ, y_{1} + r \sin θ) \\ = (0 + a \cos α, 0 + a \sin α) \\ = (a \cos α, a \sin α) \\ L e t \bar{O C} m a k i n g a n g l e w i t h X - a x i s i s \frac{π}{2} + α \\ ∴ C (x_{1} + r \cos θ, y_{1} + r \sin θ) \\ C = (0 + a \cos (\frac{π}{2} + α) 0 + a \sin α (\frac{π}{2} + α)) \\ = (- a \sin α, a \cos α) \end{array}$

$\begin{array}{l} E q u a t i o n o f \bar{A C} i s y - a \sin α = \frac{a \cos α - a \sin α}{- a \sin α - a \cos α} (x - a \cos α) \\ \Rightarrow y - a \sin α = \frac{a (\cos α - \sin α)}{- a (\cos α + \sin α)} (x - a \cos α) \\ \Rightarrow - y (\cos α + \sin α) + a \sin α \cos α + a \sin^{2} α = x (\cos α - \sin α) - a \cos^{2} α + a \cos α \sin α \\ \Rightarrow y (\cos α + \sin α) + x (\cos α - \sin α) = a \\ \Rightarrow y (\cos α + \sin α) - x (\sin α - \cos α) = a \end{array}$