Question

$A(3,2)$ and $B(-2,1)$ are two vertices of a $\Delta ABC$, whose centroid $G$ has coordinates $(\frac{5}{2},-\frac{1}{3})$. Find the coordinates of the third vertex $C$ of the triangle.

Answer 1

Let $A(3,2)$,$B(-2,1)$ and $C(x,y)$ be the coordintes of a triangle.

Centroid$(\frac{5}{2},\frac{-1}{3})=(\frac{3-2+x}{3},\frac{2+1+y}{3})$

Equating the $x-$co-ordiantes and $y-$co-ordinates we get

$\frac{1+x}{3}=\frac{5}{2}$

$2+2x=15$

or $x=\frac{13}{2}$

And $\frac{3+y}{3}=\frac{-1}{3}$

$y+3=-1$ or $y=-4$

$\therefore$ the coordinate $C$ is $(\frac{13}{2},-4)$