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