Question

If $A(6,-1)$ and $B(3,5)$ are two vertices of a triangle, find the equation of the locus of the third vertex $C$, if the centroid $G$ lies on the locus $3x^2-y^2=1$.

Answer 1

Given:

$A=(6,-1)$

$B=(3,5)$

Let $C=(h,k)$

$\therefore$ Centroid, $G=(\frac{6+3+h}{3},\frac{5-1+k}{3})=(\frac{9+h}{3},\frac{4+k}{3})$

Given, $G$ lies on the locus $3x^2-y^2=1$

$\therefore 3{\left(\frac{9+h}{3}\right)}^2-{\left(\frac{4+k}{3}\right)}^2=1$

$\Rightarrow 3h^2+54h+243-k^2-8k-16=9$

$\Rightarrow 3h^2-k^2+54h-8k+218=0$

Eq of the locus of vertex C:

$\therefore 3x^2-y^2+54x-8y+218=0$