SATHEE: Complex Numbers 4 Question 10

Complex Numbers 4 Question 10

10. $A B C D$ is a rhombus. Its diagonals $A C$ and $B D$ intersect at the point $M$ and satisfy $B D = 2 A C$ . If the points $D$ and $M$ represent the complex numbers $1 + i$ and $2 - i$ respectively, then $A$ represents the complex number …or…

(1993, 2M)

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Answer:

Correct Answer: 10. $3 - \frac{i}{2}$ or $1 - \frac{3 i}{2}$

Solution:

Given, $D = (1 + i), M = (2 - i)$

and diagonals of a rhombus bisect each other.

Let $B \equiv (a + i b)$ , therefore

$\begin{aligned} \frac{a + 1}{2} = 2, \frac{b + 1}{2} & = - 1 \\ \Rightarrow & a + 1 = 4, b + 1 & = - 2 \Rightarrow a = 3, b = - 3 \\ \Rightarrow & B & \equiv (3 - 3 i) \end{aligned}$

Again, $D M = \sqrt{(2 - 1)^{2} + (- 1 - 1)^{2}} = \sqrt{1 + 4} = \sqrt{5}$

But $B D = 2 D M \Rightarrow B D = 2 \sqrt{5}$

and $2 A C = B D \Rightarrow 2 A C = 2 \sqrt{5}$

$\Rightarrow A C = \sqrt{5}$ and $A C = 2 A M$

$\Rightarrow \sqrt{5} = 2 A M \Rightarrow A M = \frac{\sqrt{5}}{2}$

Now, let coordinate of $A$ be $(x + i y)$ .

But in a rhombus $A D = A B$ , therefore we have

$A D^{2} = A B^{2}$

$\Rightarrow (x - 1)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y + 3)^{2}$

$\Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = x^{2} + 9 - 6 x + y^{2} + 9 + 6 y$

$\Rightarrow 4 x - 8 y = 16$

$\Rightarrow x - 2 y = 4$

$\Rightarrow x = 2 y + 4$

Again, $A M = \frac{\sqrt{5}}{2} \Rightarrow A M^{2} = \frac{5}{4}$

$\Rightarrow (x - 2)^{2} + (y + 1)^{2} = \frac{5}{4}$

$\Rightarrow (2 y + 2)^{2} + (y + 1)^{2} = \frac{5}{4}$ [from Eq. (i)]

$\Rightarrow 5 y^{2} + 10 y + 5 = \frac{5}{4}$

$\Rightarrow 20 y^{2} + 40 y + 15 = 0$

$\Rightarrow 4 y^{2} + 8 y + 3 = 0$

$\Rightarrow (2 y + 1) (2 y + 3) = 0$

$\Rightarrow 2 y + 1 = 0, 2 y + 3 = 0$

$\Rightarrow y = - \frac{1}{2}, y = - \frac{3}{2}$

On putting these values in Eq. (i), we get

$\begin{aligned} x = 2 - \frac{1}{2} + 4, x = 2 - \frac{3}{2} + 4 \\ \Rightarrow & x = 3, x = 1 \end{aligned}$

Therefore, $A$ is either $3 - \frac{i}{2}$ or $1 - \frac{3 i}{2}$ .

Alternate Solution

Since, $M$ is the centre of rhombus.

$∴$ By rotating $D$ about $M$ through an angle of $\pm π / 2$ , we get possible position of $A$ .

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

$\begin{aligned} \Rightarrow z_{3} & = (2 - i) \pm \frac{1}{2} i (2 i - 1) = (2 - i) \pm \frac{1}{2} (- 2 - i) \\ = \frac{(4 - 2 i - 2 - i)}{2}, \frac{4 - 2 i + 2 + i}{2} = 1 - \frac{3}{2} i, 3 - \frac{i}{2} \end{aligned}$

$∴ A$ is either $1 - \frac{3}{2} i$ or $3 - \frac{i}{2}$ .

Complex Numbers 4 Question 10

10. $A B C D$ is a rhombus. Its diagonals $A C$ and $B D$ intersect at the point $M$ and satisfy $B D = 2 A C$ . If the points $D$ and $M$ represent the complex numbers $1 + i$ and $2 - i$ respectively, then $A$ represents the complex number …or…

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Complex Numbers 4 Question 10

10. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD=2AC. If the points D and M represent the complex numbers 1+i and 2−i respectively, then A represents the complex number …or…

Answer:

Alternate Solution

Engineering Preparation

Medical Preparation

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Syllabus

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10. $A B C D$ is a rhombus. Its diagonals $A C$ and $B D$ intersect at the point $M$ and satisfy $B D = 2 A C$ . If the points $D$ and $M$ represent the complex numbers $1 + i$ and $2 - i$ respectively, then $A$ represents the complex number …or…