Complex Numbers
Triangle Geometry in the Complex Plane
nta_pyq_2025_apr
Grade 11
Question:
Let $O$ be the origin, $A = z_1 = \sqrt{3}+2\sqrt{2}\,i$, and $B = z_2$ such that $\sqrt{3}|z_2|=|z_1|$ and $\arg(z_2)=\arg(z_1)+\dfrac{\pi}{6}$. Which of the following is true?
Area of $\triangle ABO = \dfrac{11}{\sqrt{3}}$
$\triangle ABO$ is an obtuse angled isosceles triangle
Area of $\triangle ABO = \dfrac{11}{4}$
$\triangle ABO$ is a scalene triangle
Step-by-Step Solution
Key Concept: Compute $|OA|$, $|OB|$, and $|AB|$ explicitly to classify the triangle by side lengths, then find area using $\frac{1}{2}|OA||OB|\sin(\angle AOB)$.
$|z_1|=\sqrt{3+8}=\sqrt{11}$, so $|z_2|=\sqrt{11/3}$.
$|OA|=\sqrt{11}$, $|OB|=\sqrt{11/3}$.
$|AB|^2 = |z_1|^2+|z_2|^2-2|z_1||z_2|\cos(\pi/6) = 11+\tfrac{11}{3}-2\sqrt{11}\cdot\sqrt{\tfrac{11}{3}}\cdot\tfrac{\sqrt{3}}{2} = \tfrac{44}{3}-11 = \tfrac{11}{3}$.
So $|OB|=|AB|=\sqrt{11/3}$: **isosceles** ($OB=AB$).
Since $|OA|^2=11 > \tfrac{11}{3}+\tfrac{11}{3}=\tfrac{22}{3}$, the angle at $B$ is obtuse: **obtuse isosceles**.
Area $= \dfrac{1}{2}|OA||OB|\sin\dfrac{\pi}{6} = \dfrac{1}{2}\cdot\sqrt{11}\cdot\sqrt{\dfrac{11}{3}}\cdot\dfrac{1}{2} = \dfrac{11}{4\sqrt{3}}$.
Correct Answer: 2