If the point left(alpha, frac{7 sqrt{3}}{3}ri | vedclass

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(A) The equation of the line is $x \cos 	heta + y \sin 	heta = 7$.The $x$-intercept is obtained by setting $y=0$,which gives $x = \frac{7}{\cos 	he

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$7$

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(A) The equation of the line is $x \cos 	heta + y \sin 	heta = 7$.The $x$-intercept is obtained by setting $y=0$,which gives $x = \frac{7}{\cos 	heta}$. So,the point $A$ is $\left(\frac{7}{\cos 	heta}, 0ight)$.The $y$-intercept is obtained by setting $x=0$,which gives $y = \frac{7}{\sin 	heta}$. So,the point $B$ is $\left(0, \frac{7}{\sin 	heta}ight)$.Let $M(h, k)$ be the mid-point of the line segment $AB$. Then:$h = \frac{1}{2} \left(\frac{7}{\cos 	heta} + 0ight) = \frac{7}{2 \cos 	heta} \implies \cos 	heta = \frac{7}{2h}$$k = \frac{1}{2} \left(0 + \frac{7}{\sin 	heta}ight) = \frac{7}{2 \sin 	heta} \implies \sin 	heta = \frac{7}{2k}$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\left(\frac{7}{2k}ight)^2 + \left(\frac{7}{2h}ight)^2 = 1$,which is the locus of the mid-point.The point $\left(\alpha, \frac{7 \sqrt{3}}{3}ight)$ lies on this locus,so $h = \alpha$ and $k = \frac{7 \sqrt{3}}{3}$.Substituting $k$ into the expression for $\sin 	heta$:$\sin 	heta = \frac{7}{2 \left(\frac{7 \sqrt{3}}{3}ight)} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$.Since $	heta \in \left(0, \frac{\pi}{2}ight)$,$	heta = \frac{\pi}{3}$.Then $\cos 	heta = \cos \frac{\pi}{3} = \frac{1}{2}$.Using $h = \alpha = \frac{7}{2 \cos 	heta} = \frac{7}{2 \left(\frac{1}{2}ight)} = 7$.

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7, \theta \in \left(0, \frac{\pi}{2}\right)$ between the coordinate axes,then $\alpha$ is equal to

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