Question

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

Answer 1

P is the centroid which is $\equiv (\frac{1+6+\frac{3}{2}}{3},\frac{0+2+6}{3})$
$P\equiv (\frac{17}{6},\frac{8}{3})$
$Q\equiv (-\frac{7}{6},-\frac{1}{3})$
$\therefore PQ=\sqrt{(4{)}^2+(3{)}^2}=5$