Question

If $A(\cos\alpha,\sin\alpha),\mspace{6mu} B(\sin\alpha, - \cos\alpha),C(1,\ 2)$ are the vertices of a $\Delta ABC$, then as $\alpha$varies, the locus of its centroid is.

• A. $x^{2} + y^{2} - 2x - 4y + 1 = 0$
• B. $3(x^{2} + y^{2}) - 2x - 4y + 1 = 0$
• C. $x^{2} + y^{2} - 2x - 4y + 3 = 0$
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

Let $( h , k )$ be the centroid of the triangle, then

$h = \frac { \cos \alpha + \sin \alpha + 1 } { 3 }$ and $k = \frac { \sin \alpha - \cos \alpha + 2 } { 3 }$

$\Rightarrow 3 h - 1 = \cos \alpha + \sin \alpha$ and $3 k - 2 = \sin \alpha - \cos \alpha$

$\Rightarrow ( 3 h - 1 ) ^ { 2 } + ( 3 k - 2 ) ^ { 2 } = 2$,

(squaring and adding)

$\Rightarrow 9 \left( h ^ { 2 } + k ^ { 2 } \right) - 6 h - 12 k + 3 = 0$

$\Rightarrow 3 \left( h ^ { 2 } + k ^ { 2 } \right) - 2 h - 4 k + 1 = 0$

$\therefore$ Locus of $( h , k )$is $3 \left( x ^ { 2 } + y ^ { 2 } \right) - 2 x - 4 y + 1 = 0$.