Question

If the points P(6, 2) and Q(– 2, 1) and R are the vertices of a ∆PQR and R is the point on the locus y = x² – 3x + 4, then find the equation of the locus of centroid of ∆PQR

Answer 1

P(6, 2), Q(– 2, 1).

Let R = (a, b) be a point on y = x² – 3x + 4.

Centroid of ∆PQR is `((6 - 2 + "a")/3, (2 + 1 + "b")/3)`

(i.e) `((4 + "a")/3, (3 + "b")/3)` = (h, k)

`(4 + "a")/3` = h

⇒ a = 3h – 4

`(3 + "b")/3` = k

⇒ b = 3k – 3

But (a, b) is a point on y = x² – 3x + 4

b = a² – 3a + 4

(i.e) 3k – 3 = (3h – 4)² – 3(3h – 4) + 4

(i.e) 3k – 3 = 9h² + 16 – 24h – 9h + 12 + 4

⇒ 9h² – 24h – 9h + 32 – 3k + 3 = 0

(i.e) 9h² – 33h – 3k + 35 = 0,

Locus of (h, k) is 9x² – 33x – 3y + 35 = 0