The coordinates of three points A(-4, 0),B(2, | vedclass

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(D) In an isosceles trapezium $ABCD$,the non-parallel sides are equal,i.e.,$AD = BC$. Since $B(2, 1)$ and $C(3, 1)$ lie on the line $y = 1$,the side $

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$(9, 0)$

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(D) In an isosceles trapezium $ABCD$,the non-parallel sides are equal,i.e.,$AD = BC$. Since $B(2, 1)$ and $C(3, 1)$ lie on the line $y = 1$,the side $BC$ is parallel to the $x$-axis. For $ABCD$ to be an isosceles trapezium,$AD$ must be parallel to $BC$,so $D$ must also lie on the $x$-axis (since $A$ is at $y = 0$). Let $D$ be $(x, 0)$. The length $BC = \sqrt{(3-2)^2 + (1-1)^2} = 1$. The length $AD = \sqrt{(x - (-4))^2 + (0-0)^2} = |x + 4|$. For an isosceles trapezium,the projection of the non-parallel sides on the parallel base must be equal. The $x$-coordinates of $A$ and $D$ are $-4$ and $x$. The $x$-coordinates of $B$ and $C$ are $2$ and $3$. By symmetry,the midpoint of $AD$ must align with the midpoint of $BC$. Midpoint of $BC = (\frac{2+3}{2}, 1) = (2.5, 1)$. Midpoint of $AD = (\frac{-4+x}{2}, 0) = (2.5, 0)$. $-4 + x = 5 \implies x = 9$. Thus,the coordinates of $D$ are $(9, 0)$.

The coordinates of three points $A(-4, 0)$,$B(2, 1)$,and $C(3, 1)$ determine the vertices of an isosceles trapezium $ABCD$. The coordinates of the vertex $D$ are:

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