A rectangle ABCD has its side parallel to the | vedclass

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(D) Let the vertices be $A(a, 1)$, $B(1, b)$, and $D(-1, d)$. Since $AD$ is parallel to $y=2x$, the slope of $AD$ is $2$. Thus, $\frac{d-1}{-1-a} = 2$

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None of these

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(D) Let the vertices be $A(a, 1)$, $B(1, b)$, and $D(-1, d)$. Since $AD$ is parallel to $y=2x$, the slope of $AD$ is $2$. Thus, $\frac{d-1}{-1-a} = 2$ $\Rightarrow d-1 = -2-2a$ $\Rightarrow d = -1-2a$. Since $AB$ is perpendicular to $AD$, the slope of $AB$ is $-\frac{1}{2}$. Thus, $\frac{b-1}{1-a} = -\frac{1}{2}$ $\Rightarrow 2b-2 = a-1$ $\Rightarrow b = \frac{a+1}{2}$. Since $ABCD$ is a rectangle, the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$. Midpoint of $BD = (\frac{1-1}{2}, \frac{b+d}{2}) = (0, \frac{b+d}{2})$. Midpoint of $AC = (\frac{a+x_c}{2}, \frac{1+y_c}{2})$. Equating these, we get $x_c = -a$ and $y_c = b+d-1 = \frac{a+1}{2} - 1 - 2a - 1 = \frac{a+1-2-4a-2}{2} = \frac{-3a-3}{2}$. Thus, $C$ is $(-a, \frac{-3(a+1)}{2})$. Checking the options, none of the given coordinates match this form for any real $a$.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

$A$ rectangle $ABCD$ has its side parallel to the line $y=2x$ and vertices $A, B, D$ are on lines $y=1, x=1$ and $x=-1$ respectively. The coordinate of $C$ can be

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