Question

\[ABCD\] is a rhombus. Its diagonals \[AC\] and \[BD\] intersect at the point \[M\] and satisfy \[BD=2AC\]. If the points \[D\] and \[M\] represents the complex numbers \[1+i\] and \[2-i\] respectively, then \[A\] represents the complex number
A. $3-\dfrac{1}{2}i$ or $1-\dfrac{3}{2}i$
B. $\dfrac{3}{2}-i$ or $\dfrac{1}{2}-3i$
C. $\dfrac{1}{2}-i$ or $1-\dfrac{1}{2}i$
D. None of these

Answer 1

Hint: In this question, we have to find the vertex of a rhombus. A required vertex is a complex number. By using the given diagonals and the midpoint, the required complex vertex is calculated.

Formula Used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$

Complete step by step solution: Given that,
The diagonals of the rhombus \[ABCD\] are related by \[BD=2AC\text{ }...(1)\]
Where \[AC\] and \[BD\] are the diagonals and \[M\] is the intersecting point of these diagonals.
So, \[M\] be the midpoint of the diagonals.
Then, we can write
\[\begin{align}
  & BD=BM+DM \\
 & \because DM=BM \\
 & BD=2DM\text{ }...(2) \\
\end{align}\]
And
\[\begin{align}
  & AC=AM+CM \\
 & \because AM=CM \\
 & AC=2AM\text{ }...(3) \\
\end{align}\]
From (1), (2), and (3), we can write
\[\begin{align}
  & 2DM=2(2AM) \\
 & DM=2AM \\
\end{align}\]
Then, their magnitudes are
$\begin{align}
  & \left| DM \right|=2\left| AM \right| \\
 & \Rightarrow \left| (2-i)-(1+i) \right|=2\left| (x+iy)-(2-i) \right| \\
 & \Rightarrow \left| 1-2i \right|=2\left| (x-2)+i(y+1) \right| \\
\end{align}$
But we have $\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
So,
$\begin{align}
  & \sqrt{{{1}^{2}}+{{(-2)}^{2}}}=2\sqrt{{{(x-2)}^{2}}+{{(y+1)}^{2}}} \\
 & \sqrt{5}=2\sqrt{{{(x-2)}^{2}}+{{(y+1)}^{2}}} \\
\end{align}$
Squaring on both sides, we get
$5=4\left[ {{(x-2)}^{2}}+{{(y+1)}^{2}} \right]\text{ }...(5)$
The slope of $DM$ is
$\begin{align}
  & {{m}_{1}}=\dfrac{{{y}_{2}}-{{y}_{2}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \text{ }=\dfrac{-1-1}{2-1} \\
 & \text{ }=-2 \\
\end{align}$
Since the diagonals bisect each other, they are perpendicular to each other. I.e., $DM\bot AM$
$\begin{align}
  & {{m}_{1}}\times {{m}_{2}}=-1 \\
 & \Rightarrow -2\times \dfrac{y+1}{x-2}=-1 \\
 & \Rightarrow 2(y+1)=(x-2)\text{ }...(6) \\
\end{align}$
From (5) and (6), we get
$\begin{align}
  & 5=4\left[ {{(2(y+1))}^{2}}+{{(y+1)}^{2}} \right] \\
 & \Rightarrow 5=4\left[ 4{{(y+1)}^{2}}+{{(y+1)}^{2}} \right] \\
 & \Rightarrow 5=4\left[ 5{{(y+1)}^{2}} \right] \\
 & \Rightarrow {{(y+1)}^{2}}=\dfrac{1}{4} \\
 & \Rightarrow y+1=\pm \dfrac{1}{2} \\
 & \Rightarrow y=-\dfrac{1}{2};-\dfrac{3}{2} \\
\end{align}$
Then,
\[\begin{align}
  & y=-\dfrac{1}{2}; \\
 & x-2=2(y+1) \\
 & \Rightarrow x-2=2(-\dfrac{1}{2}+1) \\
 & \Rightarrow x=3 \\
\end{align}\]
\[\begin{align}
  & y=-\dfrac{3}{2}; \\
 & x-2=2(y+1) \\
 & \Rightarrow x-2=2(-\dfrac{3}{2}+1) \\
 & \Rightarrow x-2=-1 \\
 & \Rightarrow x=1 \\
\end{align}\]
Therefore, the required complex numbers are $(3-\dfrac{1}{2}i)$ or $(1-\dfrac{3}{2}i)$

Option ‘A’ is correct

Note: Here we need to remember that, the diagonals are perpendicular to each other. So, we can able to find the required vertex.