First slide

Pair of straight lines

Question

The vertices of a triangle are $(pq, 1 / (pq)), (qr, 1 / (qr)) and (rq, 1 / (rq)),$ where p,q and r are the roots of the equation $y^{3} - 3 y^{2} + 6 y + 1 = 0$ . The coordinates of its centroid are

Moderate

A

(1,2)
B

(2,-1)
C

(1,-1)
D

(2,3)

Solution

p,q,r are the roots of equation $y^{3} - 3 y^{2} + 6 y + 1 = 0$ . So, p+q+r=3,pq+qr+rp=6, and pqr=-1. Now, the centroid of the triangle is
$(\frac{pq + qr + rp}{3}, \frac{\frac{1}{pq} + \frac{1}{qr} + \frac{1}{rp}}{3})$
i.e., $(\frac{pq + qr + rp}{3}, \frac{p + q + r}{3 pqr}) \equiv (\frac{6}{3}, \frac{3}{- 3}) or (2, - 1)$

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