Discussion
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
Answer: B
The area is \(9\sqrt{3}\) square units.
The two vertices are \(-3 - 3i\) and \(3 - 3i\), the points (-3, -3) and (3, -3). They lie on the same horizontal line, so the side is 6.
\[ \begin{aligned} \text{Area} &= \frac{\sqrt{3}}{4}a^{2} \\ &= \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \end{aligned} \]
18 is \(\frac{1}{2} \times 6 \times 6\), the area of a right isosceles triangle with legs 6, not of an equilateral triangle of side 6. The height of the equilateral triangle is \(3\sqrt{3}\), and \(\frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\) confirms the answer.
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