Question

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

• A. $3-\frac{i}{2}$ or $1-\frac{3}{2}i$
• B. $3+\frac{i}{2}$ or $1+\frac{3}{2}i$
• C. $3-i$ or $1-3i$
• D. None of these

Answer 1

The correct option is A $3-\frac{i}{2}$ or $1-\frac{3}{2}i$

Let $ABCD$ be the rhombus and $M$ be the point of intersection of the diagonals $AC$ and $BD$

Let point $D$ be $z_1=1+i$ and point $M$ be $z_2=2-i$

Also, let point $A$ be $z_3$

Then, $z_2-z_1=1-2i$ and $|z_2-z_1|=\sqrt{5}=MD$

As given, $AC=\frac{1}{2}BD\Rightarrow AM=\frac{1}{2}DM\Rightarrow AM=\frac{\sqrt{5}}{2}$

$\Rightarrow AD=|z_3-z_1|=\sqrt{{DM}^2+{AM}^2}=\sqrt{{\left(\sqrt{5}\right)}^2+{\left(\frac{\sqrt{5}}{2}\right)}^2}=\frac{5}{2}$

Therefore, in $△AMD$,

$\cos \theta =\frac{\sqrt{5}}{\frac{5}{2}}=\frac{2}{\sqrt{5}}$ and $\sin \theta =\frac{\frac{\sqrt{5}}{2}}{\frac{5}{2}}=\frac{1}{\sqrt{5}}$

Now, by rotation of complex numbers we know that $\frac{z_3-z_1}{z_2-z_1}=\frac{|z_3-z_1|}{|z_2-z_1|}{e}^{i\theta }$

(anticlockwise rotation)

$\Rightarrow \frac{z_3-(1+i)}{1-2i}=\frac{\frac{5}{2}}{\sqrt{5}}(\cos \theta +i\sin \theta )$

$\Rightarrow \frac{z_3-(1+i)}{1-2i}=1+\frac{i}{2}$ (using values of $\cos \theta$ and $\sin \theta )$

$\Rightarrow z_3=\frac{2+i}{2}(1-2i)+(1+i)\Rightarrow z_3=3-\frac{i}{2}$

Similarly , taking clockwise rotation we get another possible position of possible position of $A$ as

$\frac{z_3-z_1}{z_2-z_1}=\frac{|z_3-z_1|}{|z_2-z_1|}{e}^{-i\theta }$

$\Rightarrow z_3=\left(\frac{1-i}{2}\right)(1-2i)+(1+i)\Rightarrow z_3=1-\frac{3}{2}i$

So, $A$ represents the complex numbers $3-\frac{i}{2}$ or $1-\frac{3}{2}i$