Question

If the vertices of triangle  ABC in argand plane are the roots of equation $${z^3} + i{z^2} + 2i = 0$$, then which of the following is (are) correct[Note: where $${i^2} = - 1$$]

A
Area of triangle ABC is 2 sq. units
B
Circumradius of triangle ABC is 52
C
Triangle ABC is isosceles
D
Inradius of triangle ABC is 512

Solution

The correct option is A Area of triangle ABC is 2 sq. unitsGiven cubic equation is  $$z^3 + iz^2 + 2i = 0$$Now put $$z = i$$ in LHS  $$i^3 + i\times i^2 + 2i$$$$= i\times i^2 + i\times i^2 + 2i$$$$= -i -i +2i$$                  (since$$i^2 = -1$$)$$= -2i + 2i$$$$= 0$$= RHSSo $$z =i$$ is a factor of the given equation.Now when we divide  $$(z^3 + iz^2 + 2i)$$ by $$(z-i)$$, we get other two factors $$(1-i)$$ and $$(-1-i)$$.So three roots are $$i , 1-i, -1-i.$$Now vertices of triangle are $$(0, 1), (1, -1), (-1, -1)$$.Now are of the triangle is defined as $$\dfrac{1}{2}\times base \times height$$$$= \dfrac{1}{2}((1+1) (1+1))$$$$= \dfrac{1}{2}(2)(2)$$$$= 2$$ sq. unitsArea of the triangle is $$2$$ sq. unitsMaths

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