Verified Solution • 2025–26 Session moderate

QJEEM25Mafd90: JEE Main 2025 Mathematics - Class 11 JEEM25Mafd90 Solution: Let be the origin, the point be , the point be such that and .…

Let OO be the origin, the point AA be z1=3+22iz_1=\sqrt{3}+2 \sqrt{2} i , the point B(z2)B\left(z_2\right) be such that 3∣z2∣=∣z1∣\sqrt{3}\left|z_2\right|=\left|z_1\right| and arg⁡(z2)=arg⁡(z1)+π6\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6} . Then

Multiple Choice Options

  • A area of triangle ABO is 114\frac{11}{4}
  • B area of triangle ABO is 113\frac{11}{\sqrt{3}}
  • C ABO is a scalene triangle
  • D ABO is an obtuse angled isosceles triangle

Step-by-Step Verified Solution

z1=3+22i&∣z2∣∣z1∣=13z_1=\sqrt{3}+2 \sqrt{2} i \quad \& \frac{\left|z_2\right|}{\left|z_1\right|}=\frac{1}{\sqrt{3}}

given arg⁡(z2z1)=π6\arg \left(\frac{z_2}{z_1}\right)=\frac{\pi}{6}

amp;z2=∣z2∣∣z1∣⋅z1ei(π6)amp;z2=13⋅(3+22i)(3+i)2amp;z2=(3−22)+i(26+3)23\begin{aligned}& z_2=\frac{\left|z_2\right|}{\left|z_1\right|} \cdot z_1 \mathrm{e}^{i\left(\frac{\pi}{6}\right)} \\& z_2=\frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3}+2 \sqrt{2} i)(\sqrt{3}+i)}{2} \\& z_2=\frac{(3-2 \sqrt{2})+i(2 \sqrt{6}+\sqrt{3})}{2 \sqrt{3}}\end{aligned}

Now,

z1−z2=(3+22)+i(26−3)23z_1-z_2=\frac{(3+2 \sqrt{2})+i(2 \sqrt{6}-\sqrt{3})}{2 \sqrt{3}}

∣z1−z2∣=∣z2∣⇒ΔABO\left|z_1-z_2\right|=\left|z_2\right| \Rightarrow \Delta \mathrm{ABO} is isosceles with angles π6,π6&2π3\frac{\pi}{6}, \frac{\pi}{6} \& \frac{2 \pi}{3}

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