ABCD is a rhombus. Its diagonals AC and BD in | vedclass

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(A) In a rhombus,diagonals bisect each other at right angles. Thus,$M$ is the midpoint of $AC$ and $BD$,and $AC \perp BD$.Given $BD = 2AC$,we have $2D

Vedclass · Accepted Answer

$3 - \frac{1}{2}i$ or $1 - \frac{3}{2}i$

Vedclass · Answer

(A) In a rhombus,diagonals bisect each other at right angles. Thus,$M$ is the midpoint of $AC$ and $BD$,and $AC \perp BD$.Given $BD = 2AC$,we have $2DM = 2(2AM)$,which simplifies to $DM = 2AM$.The complex number for $D$ is $1 + i$ (point $(1, 1)$) and for $M$ is $2 - i$ (point $(2, -1)$).The vector $\vec{MD} = (1 - 2, 1 - (-1)) = (-1, 2)$.Since $AC \perp BD$,the vector $\vec{MA}$ must be perpendicular to $\vec{MD}$.Rotating $\vec{MD}$ by $90^\circ$ gives vectors $(\pm 2, \pm 1)$.Since $DM = 2AM$,the vector $\vec{MA} = \pm \frac{1}{2} \vec{MD}_{rotated} = \pm \frac{1}{2} (2, 1) = \pm (1, \frac{1}{2})$.Thus,$A = M \pm (1, \frac{1}{2}) = (2 \pm 1, -1 \pm \frac{1}{2})$.This gives $A = (3, -1/2)$ or $A = (1, -3/2)$.In complex form,$A = 3 - \frac{1}{2}i$ or $1 - \frac{3}{2}i$.

$ABCD$ is a rhombus. Its diagonals $AC$ and $BD$ intersect at the point $M$ and satisfy $BD = 2AC$. If the points $D$ and $M$ represent the complex numbers $1 + i$ and $2 - i$ respectively,then $A$ represents the complex number

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