A rectangle R with end points of one of its s | vedclass

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(C) Let the side $AB$ have endpoints $A(1, 2)$ and $B(3, 6)$.The slope of $AB$ is $m = \frac{6-2}{3-1} = \frac{4}{2} = 2$.The equation of line $AB$ is

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$16$

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(C) Let the side $AB$ have endpoints $A(1, 2)$ and $B(3, 6)$.The slope of $AB$ is $m = \frac{6-2}{3-1} = \frac{4}{2} = 2$.The equation of line $AB$ is $y - 2 = 2(x - 1)$,which simplifies to $2x - y = 0$.The given diameter is $2x - y + 4 = 0$.Since the slopes are equal,the side $AB$ is parallel to the diameter.The distance $d$ between the parallel lines $2x - y = 0$ and $2x - y + 4 = 0$ is $d = \frac{|4 - 0|}{\sqrt{2^2 + (-1)^2}} = \frac{4}{\sqrt{5}}$.Since the diameter is the perpendicular bisector of the side $AB$ in a rectangle inscribed in a circle,the distance from the diameter to the side $AB$ is half the length of the other side $BC$.Thus,$\frac{BC}{2} = \frac{4}{\sqrt{5}}$,which gives $BC = \frac{8}{\sqrt{5}}$.The length of side $AB = \sqrt{(3-1)^2 + (6-2)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.The area of the rectangle $R = AB 	imes BC = (2\sqrt{5}) 	imes \left(\frac{8}{\sqrt{5}}ight) = 16$.

$A$ rectangle $R$ with end points of one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2x - y + 4 = 0$,then the area of $R$ is

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