$A (3, 2)$ and $B (-2, 1)$ are two vertices of a triangle ABC whose centroid $G$ has the coordinates $(\frac{5}{3}, −\frac{1}{3})$. 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 $(\frac{5}{3}, −\frac{1}{3})$.
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 $(4, -4)$.
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