CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
Circumradius of triangle ABC is 52
loader
C
Triangle ABC is isosceles
loader
D
Inradius of triangle ABC is 512
loader

Solution

The correct option is A Area of triangle ABC is 2 sq. units
Given 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$$ i^2 = -1$$)

$$= -2i + 2i$$

$$= 0 $$

= RHS

So $$z =i$$ 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. units

Area of the triangle is $$2$$ sq. units

1378604_1016007_ans_0c6ff50e288a48e1b72772734ebf7385.png

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image