PointMedium
Question
If A (cos α, sin α), B (sinα, - cosα), C (1, 2) are the vertices of a ᐃABC, then as α varies, the locus of its centroid is -
Options
A.x2 + y2 - 2x - 4y + 3 = 0
B.x2 + y2 - 2x - 4y + 1 = 0
C.3(x2 + y2) - 2x - 4y + 1 = 0
D.none of these
Solution
Let (h, k) be the centroid of triangle
3h = cos α + sin α + 1
⇒ (3h - 1) = cosα + sinα .........(1)
3k = sin α - cos α+ 2
⇒ (3k - 2) = sin α - cos α .........(2)
square & add (1) & (2)
9(x2 + y2) + 6(x - 2y) = -3
3h = cos α + sin α + 1
⇒ (3h - 1) = cosα + sinα .........(1)
3k = sin α - cos α+ 2
⇒ (3k - 2) = sin α - cos α .........(2)
square & add (1) & (2)
9(x2 + y2) + 6(x - 2y) = -3
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