If A ( cos α, sin α), B ( sin α,- cos α), C (1,2) are the vertices of a triangle ABC, then as α varies, the locus of its centroid is -

Question

If A ( cos α, sin α), B ( sin α,- cos α), C (1,2) are the vertices of a triangle ABC, then as α varies, the locus of its centroid is - (A) x2+y2-2 x-4 y+3=0 (B) x2+y2-2 x-4 y+1=0 (C) 3(x2+y2)-2 x-4 y+1=0 (D) none of these

Tardigrade Admin · Accepted Answer

Let ( h , k ) be the centroid of triangle 3 h = cos α+ sin α+1 ⇒(3 h-1)= cos α+ sin α........(1) 3 k = sin α- cos α+2 ⇒(3 k-2)= sin α- cos α........(2) square add (1) (2) 9(x2+y2)+6(x-2 y)=-3

Q. If $A (cos α, sin α), B (sin α, - cos α), C (1, 2)$ are the vertices of a $△ A BC$ , then as $α$ varies, the locus of its centroid is -

$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$

$x^{2} + y^{2} - 2 x - 4 y + 1 = 0$

$3 (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0$

none of these

Let $(h, k)$ be the centroid of triangle
$3 h = cos α + sin α + 1$
$\Rightarrow (3 h - 1) = cos α + sin α$ ........(1)
$3 k = sin α - cos α + 2$
$\Rightarrow (3 k - 2) = sin α - cos α$ ........(2)
square & add (1) & (2)
$9 (x^{2} + y^{2}) + 6 (x - 2 y) = - 3$

Q. If A(cosα,sinα),B(sinα,−cosα),C(1,2) are the vertices of a △ABC, then as α varies, the locus of its centroid is -

x2+y2−2x−4y+3=0

x2+y2−2x−4y+1=0

3(x2+y2)−2x−4y+1=0