A (3, 2) and B (−2, 1) are two vertices of a triangle ABC whose

Question:

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

Solution:

We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be $(x, y)$.

The co-ordinates of other two vertices are $A(3,2)$ and $C(-2,1)$

The co-ordinate of the centroid is $\left(\frac{5}{3},-\frac{1}{3}\right)$

We know that the co-ordinates of the centroid of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is-

$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

So,

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

Compare individual terms on both the sides-

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

So,

$x=4$

Similarly,

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

So,

$y=-4$

So the co-ordinate of third vertex $(4,-4)$