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If a vertex of a triangle is $(1,1)$ and the midpoints of two sides of the triangle through this vertex are $(-1,2)$ and $(3,2)$, then the centroid of the triangle is
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The correct answer is:
$\left(1, \frac{7}{3}\right)$
Midpoint of $\mathrm{AB}=(-1,2)$
$\Rightarrow\left(\frac{\mathrm{x}_{2}+1}{2}, \frac{\mathrm{y}_{2}+1}{2}\right)=(-1,2)$
$\Rightarrow \frac{\mathrm{x}_{2}+1}{2}=-1 ; \frac{\mathrm{y}_{2}+1}{2}=2$
$\Rightarrow \mathrm{x}_{2}+1=-2 ; \mathrm{y}_{2}+1=4$
$\Rightarrow \mathrm{x}_{2}=-3, \mathrm{y}_{2}=3$
Midpoint of $\mathrm{AC}=(3,2)$
$\Rightarrow\left(\frac{\mathrm{x}_{1}+1}{2}, \frac{\mathrm{y}_{1}+1}{2}\right)=(3,2)$
$\Rightarrow \mathrm{x}_{1}+1=6 ; \mathrm{y}_{1}+1=4$
$\Rightarrow \mathrm{x}_{1}=5, \mathrm{y}_{1}=3$
So, vertices of triangle $\mathrm{ABC}$ are $(1,1),(-3,3),(5,3)$
So, centroid $=\left(\frac{1-3+5}{3}, \frac{1+3+3}{3}\right)=\left(\frac{3}{3}, \frac{7}{3}\right)=\left(1, \frac{7}{3}\right)$
$\Rightarrow\left(\frac{\mathrm{x}_{2}+1}{2}, \frac{\mathrm{y}_{2}+1}{2}\right)=(-1,2)$
$\Rightarrow \frac{\mathrm{x}_{2}+1}{2}=-1 ; \frac{\mathrm{y}_{2}+1}{2}=2$
$\Rightarrow \mathrm{x}_{2}+1=-2 ; \mathrm{y}_{2}+1=4$
$\Rightarrow \mathrm{x}_{2}=-3, \mathrm{y}_{2}=3$
Midpoint of $\mathrm{AC}=(3,2)$
$\Rightarrow\left(\frac{\mathrm{x}_{1}+1}{2}, \frac{\mathrm{y}_{1}+1}{2}\right)=(3,2)$
$\Rightarrow \mathrm{x}_{1}+1=6 ; \mathrm{y}_{1}+1=4$
$\Rightarrow \mathrm{x}_{1}=5, \mathrm{y}_{1}=3$
So, vertices of triangle $\mathrm{ABC}$ are $(1,1),(-3,3),(5,3)$
So, centroid $=\left(\frac{1-3+5}{3}, \frac{1+3+3}{3}\right)=\left(\frac{3}{3}, \frac{7}{3}\right)=\left(1, \frac{7}{3}\right)$
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