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प्रश्न
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
उत्तर
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
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