Let A(1,0), B(6,2) and C(frac{3}{2}, 6) be th | vedclass

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(B) If a point $P$ inside a triangle $ABC$ divides it into three triangles $APC, APB,$ and $BPC$ of equal areas,then $P$ must be the centroid of the t

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$5$

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(B) If a point $P$ inside a triangle $ABC$ divides it into three triangles $APC, APB,$ and $BPC$ of equal areas,then $P$ must be the centroid of the triangle $ABC$.The coordinates of the centroid $P(x, y)$ are given by the average of the coordinates of the vertices $A(1, 0), B(6, 2),$ and $C(\frac{3}{2}, 6)$:$x = \frac{1 + 6 + 1.5}{3} = \frac{8.5}{3} = \frac{17}{6}$$y = \frac{0 + 2 + 6}{3} = \frac{8}{3}$So,$P = (\frac{17}{6}, \frac{8}{3})$.We need to find the distance $PQ$ where $Q = (-\frac{7}{6}, -\frac{1}{3})$:$PQ = \sqrt{(\frac{17}{6} - (-\frac{7}{6}))^2 + (\frac{8}{3} - (-\frac{1}{3}))^2}$$PQ = \sqrt{(\frac{24}{6})^2 + (\frac{9}{3})^2}$$PQ = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

Let $A(1,0), B(6,2)$ and $C(\frac{3}{2}, 6)$ 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

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