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One side of a square is inclined at an acute angle $\alpha$ with the positive $x-$axis, and one of its extremities is at the origin. If the remaining three vertices of the square lie above the $x-$axis and the side of a square is $4$, then the equation of the diagonal of the square which is not passing through the origin is
$(cos\, \alpha + sin\, \alpha) x + (cos\, \alpha - sin\, \alpha) y = 4$
$(cos\, \alpha + sin\, \alpha) x - (cos\, \alpha - sin\, \alpha) y = 4$
$(cos\, \alpha - sin\, \alpha) x + (cos\, \alpha + sin\, \alpha) y = 4$
$(cos\, \alpha - sin\, \alpha) x - (cos\, \alpha + sin\, \alpha) y = 4 cos\, 2\alpha$
Solution
$\therefore \,\,\,$ slope of line $AC =$ $\frac{{\not 4\sin\, \alpha – \not 4\cos\, \alpha }}{{\not 4\cos\, \alpha + \not 4\sin\, \alpha }}$
$\therefore \,\,\,$ equation of line = $\frac{{\sin\, \alpha – \sin\, \alpha }}{{\cos\, \alpha \sin\, \alpha }}$
$y – sin\, \alpha =$ $\frac{{\sin\, \alpha – \sin\, \alpha }}{{\cos\, \alpha \sin\, \alpha }}$ $(x – 4cos\, \alpha)$
$y\, cos\, \alpha + y\, sin\, \alpha – 4 \, sin\, \alpha · cos\, \alpha = x \,sin\, \alpha – x \,cos\, \alpha – 4 \,sin^2\, \alpha – sin\, \alpha cos\, \alpha + 4\, cos^2\, \alpha$
$x (cos\, \alpha – sin\, \alpha) + y (cos\, \alpha + sin\, \alpha) = 4\Rightarrow(D)$
