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

Given:

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

To do:

We have to find the coordinates of the third vertex $C$ of the triangle.

Solution:

Let the coordinates of the third vertex be $C(x,y)$.

We know that,

Coordinates of the centroid of a triangle are $(\frac{Sum\ of\ abscissa}{3}, \frac{Sum\ of\ ordinates}{3})$

Therefore,

The coordinates of the centroid of the triangle ABC are,

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

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

On comparing, we get,

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

$5=1+x$

$x=5-1$

$x=4$

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

$-1=3+y$

$y=-1-3$

$y=-4$

The coordinates of the third vertex $C$ are $(4, -4)$.

Updated on: 10-Oct-2022

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