Let A (1 , 0), B (6 , 2) and C (3/2 , 6 ) be the vertices of

Question:

Let $A(1,0), B(6,2)$ and $C\left(\frac{3}{2}, 6\right)$ be the vertices of a triangle $\mathrm{ABC}$. If $\mathrm{P}$ is a point inside the triangle $\mathrm{ABC}$ such that the triangles APC, APB and BPC have equal areas, then the length of the line segment $\mathrm{PQ}$, where $\mathrm{Q}$ is the point $\left(-\frac{7}{6},-\frac{1}{3}\right)$, is__________.

Solution:

$P$ is centroid of the triangle $A B C$

$\Rightarrow P \equiv\left(\frac{17}{6}, \frac{8}{3}\right)$

$\Rightarrow P Q=5$

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