
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
164.4k+ views
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.
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.
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