$A(1,4),$ $B(2,3)$ are the vertices of $\triangle ABC$. Centroid of the triangle lies on the locus $3x-9y+13=0$. If the equation of the locus of the third vertex $C$ is $x-by-c=0$ then find $(b+c)$.
My Solution -
Let $C\equiv (h,k) ⇒ G\equiv \Big(\dfrac{h+3}{3},\dfrac{k+7}{3}\Big)$
As $G $ satisfies the locus $⇒$ $h+3-3(k+7)+13=0⇒h+16-3k-21=0⇒h-3k-5=0$
Replacing $(h,k)$ by $x,y⇒$ Equation of $C⇒$ $x-3y-5=0⇒b+c=\boxed{8}.$
I am not sure if I got the locus of $C$ right but if it is correct then why am I getting this answer? I mean just by plugging $G\equiv \Big(\dfrac{h+3}{3},\dfrac{k+7}{3}\Big)$ in the locus of $G$ why am I getting equation of locus of $C$? Can someone please clarify?
The question setter is telling you that as $C$ moves along the line $x-by-c=0$ then $G$ moves along the line $3x-9y+13=0$.
As you have realised, if you let $C$ be $(h,k)$ then $G$ is $\Big(\dfrac{h+3}{3},\dfrac{k+7}{3}\Big)$.
When you now substitute the coordinates of $G$ into the equation $3x-9y+13=0$ you will obtain a linear equation between $h$ and $k$, which is exactly what the equation $x-by-c=0$ is.