A square of side length a has one vertex at t | vedclass

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Question

(A) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B(a \cos(\alpha + \pi/4) \sqrt{2}, a \sin(\alpha + \pi/4) \sqrt{2})$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B(a \cos(\alpha + \pi/4) \sqrt{2}, a \sin(\alpha + \pi/4) \sqrt{2})$,and $C$. Since the side $OA$ makes an angle $\alpha$ with the $x$-axis,its length is $a$. The vertex $A$ is $(a \cos \alpha, a \sin \alpha)$. The diagonal not passing through the origin is $AC$. The slope of $OA$ is $	an \alpha$. The side $OC$ is perpendicular to $OA$,so its slope is $	an(\alpha + \pi/2) = -\cot \alpha$. The vertex $C$ is $(a \cos(\alpha + \pi/2), a \sin(\alpha + \pi/2)) = (-a \sin \alpha, a \cos \alpha)$. The diagonal $AC$ passes through $A(a \cos \alpha, a \sin \alpha)$ and $C(-a \sin \alpha, a \cos \alpha)$. The slope of $AC$ is $m = \frac{a \cos \alpha - a \sin \alpha}{-a \sin \alpha - a \cos \alpha} = \frac{\cos \alpha - \sin \alpha}{-(\sin \alpha + \cos \alpha)}$. The equation of $AC$ is $y - a \sin \alpha = \frac{\cos \alpha - \sin \alpha}{-(\sin \alpha + \cos \alpha)} (x - a \cos \alpha)$. $-y(\sin \alpha + \cos \alpha) + a \sin \alpha (\sin \alpha + \cos \alpha) = x(\cos \alpha - \sin \alpha) - a \cos \alpha (\cos \alpha - \sin \alpha)$. $x(\cos \alpha - \sin \alpha) + y(\sin \alpha + \cos \alpha) = a(\sin^2 \alpha + \sin \alpha \cos \alpha + \cos^2 \alpha - \sin \alpha \cos \alpha) = a(1) = a$. Thus,the equation is $y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a$.

$A$ square of side length $a$ has one vertex at the origin and one side along the $x$-axis. The side passing through the origin makes an angle $\alpha$ $(0 < \alpha < \pi/4)$ with the positive $x$-axis. Find the equation of the diagonal that does not pass through the origin.

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