In a rectangle ABCD,the coordinates of A and | vedclass

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(A) Given that $ABCD$ is a rectangle with vertices $A(1, 2)$ and $B(3, 6)$.The equation of one of the diameters of the circumscribing circle is $2x -

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$16$

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(A) Given that $ABCD$ is a rectangle with vertices $A(1, 2)$ and $B(3, 6)$.The equation of one of the diameters of the circumscribing circle is $2x - y + 4 = 0$.The slope of this diameter is $m_1 = 2$.The slope of the side $AB$ is $m_{AB} = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$.Since the slope of $AB$ is equal to the slope of the diameter,the side $AB$ is parallel to the diameter.The distance $d$ between the parallel lines (the diameter and the line $AB$) is the perpendicular distance from any point on $AB$ (e.g.,$A(1, 2)$) to the line $2x - y + 4 = 0$:$d = \frac{|2(1) - 1(2) + 4|}{\sqrt{2^2 + (-1)^2}} = \frac{|2 - 2 + 4|}{\sqrt{5}} = \frac{4}{\sqrt{5}}$.In a rectangle,the center of the circumscribing circle is the intersection of the diagonals. The distance from the center to the side $AB$ is half the length of the side $BC$. Thus,$BC = 2d = 2 	imes \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}}$.The length of side $AB$ is $\sqrt{(3 - 1)^2 + (6 - 2)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.The area of the rectangle is $AB 	imes BC = 2\sqrt{5} 	imes \frac{8}{\sqrt{5}} = 16$.

In a rectangle $ABCD$,the coordinates of $A$ and $B$ are $(1, 2)$ and $(3, 6)$ respectively,and some diameter of the circumscribing circle of $ABCD$ has the equation $2x - y + 4 = 0$. Then,the area of the rectangle is

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