One side of a square is inclined at an acute | vedclass

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(C) Let the square be $ABCD$ with vertex $B$ at the origin $(0,0)$. The side $BC$ makes an angle $\alpha$ with the positive $x-$axis. Since the side l

Vedclass · Accepted Answer

$(\cos\, \alpha - \sin\, \alpha) x + (\cos\, \alpha + \sin\, \alpha) y = 4$

Vedclass · Answer

(C) Let the square be $ABCD$ with vertex $B$ at the origin $(0,0)$. The side $BC$ makes an angle $\alpha$ with the positive $x-$axis. Since the side length is $4$,the coordinates of $C$ are $(4 \cos \alpha, 4 \sin \alpha)$.Since the square is in the first quadrant,the side $BA$ is perpendicular to $BC$ and makes an angle $\alpha + 90^\circ$ with the $x-$axis. Thus,the coordinates of $A$ are $(4 \cos(\alpha + 90^\circ), 4 \sin(\alpha + 90^\circ)) = (-4 \sin \alpha, 4 \cos \alpha)$.The vertex $D$ is $A + C = (4 \cos \alpha - 4 \sin \alpha, 4 \sin \alpha + 4 \cos \alpha)$.The diagonal not passing through the origin is $AD$. The slope of $AD$ is $\frac{(4 \sin \alpha + 4 \cos \alpha) - 4 \cos \alpha}{(4 \cos \alpha - 4 \sin \alpha) - (-4 \sin \alpha)} = \frac{4 \sin \alpha}{4 \cos \alpha} = 	an \alpha$.The equation of line $AD$ passing through $A(-4 \sin \alpha, 4 \cos \alpha)$ is $y - 4 \cos \alpha = 	an \alpha (x + 4 \sin \alpha)$.$y \cos \alpha - 4 \cos^2 \alpha = x \sin \alpha + 4 \sin^2 \alpha$.$x \sin \alpha - y \cos \alpha + 4(\sin^2 \alpha + \cos^2 \alpha) = 0 \Rightarrow x \sin \alpha - y \cos \alpha + 4 = 0$.However,checking the diagonal $AC$: The slope of $AC$ is $\frac{4 \sin \alpha - 4 \cos \alpha}{4 \cos \alpha - (-4 \sin \alpha)} = \frac{\sin \alpha - \cos \alpha}{\cos \alpha + \sin \alpha}$.The equation of $AC$ passing through $C(4 \cos \alpha, 4 \sin \alpha)$ is $y - 4 \sin \alpha = \frac{\sin \alpha - \cos \alpha}{\cos \alpha + \sin \alpha} (x - 4 \cos \alpha)$.$(y - 4 \sin \alpha)(\cos \alpha + \sin \alpha) = (x - 4 \cos \alpha)(\sin \alpha - \cos \alpha)$.$y(\cos \alpha + \sin \alpha) - 4 \sin \alpha \cos \alpha - 4 \sin^2 \alpha = x(\sin \alpha - \cos \alpha) - 4 \sin \alpha \cos \alpha + 4 \cos^2 \alpha$.$x(\cos \alpha - \sin \alpha) + y(\cos \alpha + \sin \alpha) = 4(\cos^2 \alpha + \sin^2 \alpha) = 4$.

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