Q48 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
If the roots of the equation $Z^3+i Z^2+2 i=0$ are the vertices of a triangle $A B C$. then that triangle $A B C$ is
- Aa right angled triangle
- Ban equilateral triangle
- Correct answerCan isosceles triangle
- Da right angled isosceles triangle
Solution
Given, $Z^3+i Z^2+2 i=0$
$\begin{aligned}
& \Rightarrow(Z-i)\left(Z^2+2 i Z-2\right)=0 \\
& Z=i \text { or } Z^2+2 i Z-2=0 \\
& \Rightarrow Z=\frac{-2 i \pm 2}{2} \Rightarrow z=-i+1,-i-1
\end{aligned}$
Let $A(0,1), B(1,-1), C(-1,-1)$ $A B=\sqrt{5} ; B C=2 ; A C=\sqrt{5}$
Since, $A B^2+A C^2 \neq B C^2$ and $A B=B C \neq B C$. So, $\triangle A B C$ is an isosceles triangle.
Let $A(0,1), B(1,-1), C(-1,-1)$ $A B=\sqrt{5} ; B C=2 ; A C=\sqrt{5}$
Since, $A B^2+A C^2 \neq B C^2$ and $A B=B C \neq B C$. So, $\triangle A B C$ is an isosceles triangle.
Hence, the correct answer is option: 'C'.