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If a vertex of traingle is $(3,3)$ and the mid-points of two sides through this vertex are $\left(2, \frac{3}{2}\right)$ and $\left(4, \frac{3}{2}\right)$, then the centroid of the triangle is given by
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The correct answer is:
$(3,1)$
Let the vertices of $\triangle A B C$ are $A(3,3), B\left(x_{1}, y_{1}\right)$ and $C\left(x_{2}, y_{2}\right)$.
Now, $\frac{x_{1}+3}{2}=2$ and $\frac{y_{1}+3}{2}=\frac{3}{2}$ $x_{1}+3=4$ and $y_{1}+3=3$
$x_{1}=1$ and $y_{1}=0$
So, $\quad B \equiv(1,0)$
and $\frac{x_{2}+3}{2}=4$ and $\frac{y_{2}+3}{2}=\frac{3}{2}$
$x_{2}+3=8$ and $y_{2}+3=3$
$x_{2}=5$ and $y_{2}=0$
$C \equiv(5,0)$
Hence, centroid $=\left(\frac{3+1+5}{3}, \frac{3+0+0}{3}\right)$
$$
=\left(\frac{9}{3}, \frac{3}{3}\right)=(3,1)
$$
Now, $\frac{x_{1}+3}{2}=2$ and $\frac{y_{1}+3}{2}=\frac{3}{2}$ $x_{1}+3=4$ and $y_{1}+3=3$
$x_{1}=1$ and $y_{1}=0$
So, $\quad B \equiv(1,0)$
and $\frac{x_{2}+3}{2}=4$ and $\frac{y_{2}+3}{2}=\frac{3}{2}$
$x_{2}+3=8$ and $y_{2}+3=3$
$x_{2}=5$ and $y_{2}=0$
$C \equiv(5,0)$
Hence, centroid $=\left(\frac{3+1+5}{3}, \frac{3+0+0}{3}\right)$
$$
=\left(\frac{9}{3}, \frac{3}{3}\right)=(3,1)
$$
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