First slide

Pair of straight lines

Question

If $α, β, γ$ are the real roots of the equation $x^{3} - 3 p x^{2} + 3 q x - 1 = 0$ , then the centroid of the triangle with vertices $(α, \frac{1}{α}), (β, \frac{1}{β}) and (γ, \frac{1}{γ})$ is at the point

Moderate

A

$(p, q)$
B

$(p / 3, q / 3)$
C

$(p + q, p - q)$
D

$(3 p, 3 q)$

Solution

The centroid of the given triangle is the point
$(\frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3}) = (\frac{3 p}{3}, \frac{α β + β γ + γ α}{3 α β γ})$
$= (p, q) [∵ α + β + γ = 3 p, α β + β γ + γ α = 3 q, α β γ = 1]$

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