Question

The two vertices \[B\] and \[D\] of a parallelogram are \[1-2i\] and \[4+2i\], if the diagonals are at right angles and \[AC=2BD\], then complex numbers representing \[A\] is
A. $\dfrac{5}{2}$
B. $3i-\dfrac{3}{2}$
C. $3i-4$
D. $3i+4$

Answer 1

Hint: In this question, we have to find the complex number representing the vertex of the given parallelogram. By applying the given condition \[AC=2BD\], we can able to find the required complex number.

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 parallelogram has two vertices as
\[B=1-2i\]
\[D=4+2i\]
For the given parallelogram, \[\overrightarrow{BD}\] is one of the diagonals.
So, we can calculate its magnitude by
$\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
$\begin{align}
  & \left| \overrightarrow{BD} \right|=\left| (4+2i)-(1-2i) \right| \\
 & \text{ }=\left| (4-1)+(2i+2i) \right| \\
 & \text{ }=\sqrt{{{(3)}^{2}}+{{(4)}^{2}}} \\
 & \text{ }=\sqrt{25} \\
 & \text{ }=5 \\
\end{align}$
So, the midpoint of the diagonal is
$\begin{align}
  & E=\left( \dfrac{1+4}{2},\dfrac{-2+2}{2} \right) \\
 & \text{ }=(\dfrac{5}{2},0) \\
\end{align}$
Since the diagonals of a parallelogram bisect each other, the point of intersection of the diagonals is $\left( \dfrac{5}{2},0 \right)$.
From the given condition,
\[\begin{align}
  & AC=2BD \\
 & \Rightarrow BD=\dfrac{1}{2}AC \\
\end{align}\]
But we know,
\[\dfrac{1}{2}AC=AE\]
Therefore,
$\left| \overrightarrow{BD} \right|=\left| \overrightarrow{AE} \right|$
So, we can write
$\begin{align}
  & \left| \overrightarrow{AE} \right|=\left| (\dfrac{5}{2}+0i)-(x+iy) \right| \\
 & \text{ }=\left| (\dfrac{5}{2}-x)-iy \right| \\
 & \text{ }=\sqrt{{{(\dfrac{5}{2}-x)}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus,
$\begin{align}
  & \left| \overrightarrow{AE} \right|=5=\sqrt{{{(\dfrac{5}{2}-x)}^{2}}+{{y}^{2}}} \\
 & \Rightarrow 25=\dfrac{25}{4}-5x+{{x}^{2}}+{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-5x=\dfrac{75}{4}\text{ }...(1) \\
\end{align}$
In order to find the complex number at vertex \[A\], substituting the given option in the obtained equation.
I.e.,
Since the option A has only one value, it is not valid for the required complex number.
Now, substituting the option B, we get
$3i-\dfrac{3}{2};x=-\dfrac{3}{2}\And y=3$
So,
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}-5x={{\left( \dfrac{-3}{2} \right)}^{2}}+{{3}^{2}}-5\left( \dfrac{-3}{2} \right) \\
 & \text{ }=\dfrac{9}{4}+9+\dfrac{15}{2} \\
 & \text{ }=\dfrac{9+36+30}{4} \\
 & \text{ }=\dfrac{75}{4} \\
\end{align}$
Thus, the required complex number is $3i-\dfrac{3}{2}$.
Option ‘B’ is correct

Note: Here we need to remember that, the point of intersection of the diagonals is the midpoint of the given diagonal. By using this, the required complex number is calculated.