You visited us 162 times! Enjoying our articles? Unlock Full Access!

Family of Straight Lines

The roots z 1...

Question

The roots $z_{1}, z_{2}, z_{3}$ of the equation $x^{3} + 3 p x^{2} + 3 q x + r = 0$ (p,q,r are complex) correspond to points A, B and C. Then triangle ABC is equilateral if

p = q^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

p^{2} = 3 q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

p^{2} = q

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

q^{2} = 3 p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $p^{2} = q$
we have $z_{1} + z_{2} + z_{3} = - 3 p, z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2} = 3 q$ and $z_{1} z_{2} z_{3} = - r$
Triangle $A (z 1), B (z_{2}),$ and $c (z_{3})$ is an equilateral triangle if and only if
$\frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} + \frac{1}{z_{1} - z_{2}} = 0$
$\Leftrightarrow (z_{3} - z_{1}) (z_{1} - z_{2}) + (z_{2} - z_{3}) (z_{2} - z_{3}) + (z_{2} - z_{3}) (z_{3} - z_{1}) = 0$
$\Leftrightarrow (z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{2} z_{3} + z_{3} z_{1} + z_{3} z_{1} + z_{1} z_{2}$
$\Leftrightarrow (z_{1} + z_{2} + z_{3})^{2} = 3 (z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2})$
$\Leftrightarrow (- 3 p)^{2} = 3 (3 q) \Leftrightarrow p^{2} = q$

Suggest Corrections

Similar questions

z_{1}

z_{2}

z^{2} + p z + q = 0

p

q

z_{1}, z_{2}

z_{1}, z_{2}, z_{3}

p, q, r

\frac{p}{| z_{2} - z_{3} |} = \frac{q}{| z_{3} - z_{1} |} = \frac{r}{| z_{1} - z_{2} |}

\frac{p^{2}}{z_{2} - z_{3}} + \frac{q^{2}}{z_{3} - z_{1}} + \frac{r^{2}}{z_{1} - z_{2}} = 0

z_{1}, z_{2}, z_{3}

\frac{p}{| z_{2} - z_{3} |} = \frac{q}{| z_{3} - z_{1} |} = \frac{r}{| z_{1} - z_{2} |}

\frac{p^{2}}{z_{2} - z_{3}} + \frac{q^{2}}{z_{3} - z_{1}} + \frac{r^{2}}{z_{1} - z_{2}} = 0

z_{1}

z_{2}

z_{3}

x^{3} + 3 a x^{2} + 3 b x + c = 0

p, q a n d r

(p \neq q, r \neq 0)

\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}

Join BYJU'S Learning Program