Question

In a complex plane, if two vertices of an equilateral triangle are at \(-3(1+i)\) and \(3(1-i)\), then the area of the triangle (in sq.units) is equal to

• A. 18
• B. \(9\sqrt{3}\)
• C. \(6\sqrt{3}\)
• D. \(3\sqrt{3}\)
• E. 9

Hint:

Notice that the imaginary parts of both vertices are \(-3i\). This means the segment is perfectly horizontal on the complex plane. The distance is simply the absolute difference of the real parts: \(|3 - (-3)| = 6\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The area of an equilateral triangle is solely determined by its side length.
We calculate the distance between the two given complex coordinates to find the side length \(s\).

Step 2: Key Formula or Approach:
1. Distance between two complex numbers \(z_1\) and \(z_2\) is \(s = |z_1 - z_2|\).
2. Area of equilateral triangle \(A = \frac{\sqrt{3}}{4}s^2\).

Step 3: Detailed Explanation:
Let \(z_1 = -3(1+i) = -3 - 3i\).
Let \(z_2 = 3(1-i) = 3 - 3i\).
Find side length \(s\):
\[ s = |(-3 - 3i) - (3 - 3i)| \]
\[ s = |-3 - 3i - 3 + 3i| \]
\[ s = |-6| = 6 \]
Calculate Area:
\[ \text{Area} = \frac{\sqrt{3}}{4} \cdot (6)^2 \]
\[ \text{Area} = \frac{\sqrt{3}}{4} \cdot 36 = 9\sqrt{3} \]

Step 4: Final Answer:
The area of the triangle is \(9\sqrt{3}\) sq.units.