Let O be the origin, the point A be \(z_1\), and the point B be \(z_2\) such that \(|z_2|=\sqrt{3}\,|z_1|\) and \(\arg(z_2)=\arg(z_1)+\frac{\pi}{6}\). Then the triangle ABO is:

(D) ABO is an obtuse angled isosceles triangle

(A) area of triangle ABO is 11/3

(C) area of triangle ABO is 11/4

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Complex Numbers

Let O be the origin, the point A be

z_{1}

, and the point B be

z_{2}

such that

∣ z_{2} ∣ = 3 ∣ z_{1} ∣

and

ar g (z_{2}) = ar g (z_{1}) + \frac{π}{6}

. Then the triangle ABO is:

Answer: (D)

From

∣ z_{2} ∣ = 3 ∣ z_{1} ∣

and

ar g z_{2} = ar g z_{1} + \frac{π}{6}

, the angle

∠ A O B = \frac{π}{6}

. Writing

O A = r, O B = 3 r

A B^{2} = r^{2} + 3 r^{2} - 23 r^{2} cos \frac{π}{6} = 4 r^{2} - 3 r^{2} = r^{2}

, so

A B = r = O A

. The triangle is isosceles, and the angle opposite the longest side

O B

\frac{2 π}{3}

; hence it is an obtuse-angled isosceles triangle.

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option