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(A, C) Let the vertices of the square be $O(0,0)$, $A(a \cos \alpha, a \sin \alpha)$, $C(a \cos(\alpha + 90^{\circ}), a \sin(\alpha + 90^{\circ})) = (

Vedclass · Accepted Answer

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

Vedclass · Answer

(A, C) Let the vertices of the square be $O(0,0)$, $A(a \cos \alpha, a \sin \alpha)$, $C(a \cos(\alpha + 90^{\circ}), a \sin(\alpha + 90^{\circ})) = (-a \sin \alpha, a \cos \alpha)$, and $B(a \cos \alpha - a \sin \alpha, a \sin \alpha + a \cos \alpha)$.The first diagonal passes through $O(0,0)$ and $B(a(\cos \alpha - \sin \alpha), a(\sin \alpha + \cos \alpha))$.The slope of this diagonal is $m_1 = \frac{a(\sin \alpha + \cos \alpha)}{a(\cos \alpha - \sin \alpha)} = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha}$.The equation is $y = m_1 x \Rightarrow y(\cos \alpha - \sin \alpha) = x(\sin \alpha + \cos \alpha)$.The second diagonal passes through $A(a \cos \alpha, a \sin \alpha)$ and $C(-a \sin \alpha, a \cos \alpha)$.The slope of this diagonal is $m_2 = \frac{a \cos \alpha - a \sin \alpha}{-a \sin \alpha - a \cos \alpha} = -\frac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}$.The equation is $y - a \sin \alpha = m_2(x - a \cos \alpha)$.$(y - a \sin \alpha)(\cos \alpha + \sin \alpha) = -(\cos \alpha - \sin \alpha)(x - a \cos \alpha)$.$y(\cos \alpha + \sin \alpha) - a \sin \alpha \cos \alpha - a \sin^2 \alpha = -x(\cos \alpha - \sin \alpha) + a \cos^2 \alpha - a \sin \alpha \cos \alpha$.$y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a(\cos^2 \alpha + \sin^2 \alpha) = a$.

$A$ square with each side of length $a$ lies above the $x$-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle $\alpha$ $(0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. Find the equations of the diagonals of the square.

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