The vertices B and D of a parallelogram are 1 | vedclass

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(B) Let the vertices be $A, B, C, D$ in order. The diagonals $AC$ and $BD$ bisect each other at $E$. The midpoint $E$ of $BD$ is $\frac{(1 - 2i) + (4

Vedclass · Accepted Answer

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

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(B) Let the vertices be $A, B, C, D$ in order. The diagonals $AC$ and $BD$ bisect each other at $E$. The midpoint $E$ of $BD$ is $\frac{(1 - 2i) + (4 + 2i)}{2} = \frac{5}{2}$. Since $AC = 2BD$,the length $AE = EC = BD = |(4 + 2i) - (1 - 2i)| = |3 + 4i| = \sqrt{3^2 + 4^2} = 5$. Since the diagonals are at right angles,the vector $\vec{EA}$ is obtained by rotating $\vec{ED}$ by $90^\circ$ (or $-90^\circ$). $\vec{ED} = (4 + 2i) - \frac{5}{2} = \frac{3}{2} + 2i$. Rotating by $90^\circ$ (multiplying by $i$): $\vec{EA} = i(\frac{3}{2} + 2i) = -2 + \frac{3}{2}i$. Thus,$A = E + \vec{EA} = \frac{5}{2} + (-2 + \frac{3}{2}i) = \frac{1}{2} + \frac{3}{2}i$. Rotating by $-90^\circ$ (multiplying by $-i$): $\vec{EA} = -i(\frac{3}{2} + 2i) = 2 - \frac{3}{2}i$. Thus,$A = E + \vec{EA} = \frac{5}{2} + (2 - \frac{3}{2}i) = \frac{9}{2} - \frac{3}{2}i$. Checking the options,option $(b)$ is $3i - \frac{3}{2} = -1.5 + 3i$. Re-evaluating the geometry,if $A$ is $z$,then $z - E = \pm i(D - E)$. $z = \frac{5}{2} \pm i(\frac{3}{2} + 2i) = \frac{5}{2} \pm (\frac{3}{2}i - 2) = (\frac{5}{2} \mp 2) \pm \frac{3}{2}i$. This gives $A = \frac{1}{2} + \frac{3}{2}i$ or $A = \frac{9}{2} - \frac{3}{2}i$. Given the provided options,there might be a typo in the question or options. Based on standard interpretation,none match exactly,but we provide the derivation.

The vertices $B$ and $D$ of a parallelogram are $1 - 2i$ and $4 + 2i$. If the diagonals are at right angles and $AC = 2BD$,the complex number representing $A$ is

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