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 triangle $APC,APBandBPC$ have equal areas, then the length of the line the segment $PQ$, where $Q$ is the point $\left(\frac{-7}{6},\frac{-1}{3}\right)$,is

Answer 1

Finding the length of line segment $PQ$

Step 1: Determine the value of $P$

Given the vertices of a triangle $ABC$are $A(1,0)$, $B(6,2)$ and $C\left(\frac{3}{2},6\right)$.

$P$ is a point inside the given triangle such that the triangles $APC,APB$and $BPC$ have equal area, therefore the point $P$ will be the centroid of a triangle $ABC$.

Point $Q$ has coordinates $\left(-\frac{7}{6},-\frac{1}{3}\right)$

The centroid of a triangle is calculated by adding the vertices of a triangle and dividing by $3$. Hence,

$$\begin{gathered} P=\left(\frac{1+6+{\displaystyle \frac{3}{2}}}{3},\frac{0+2+6}{3}\right) \\ P=\left(\frac{17}{6},\frac{8}{3}\right) \end{gathered}$$

Step 2: Determine the Length of $PQ$

Now the length of a line segment $PQ$ will be: $\sqrt{(x_2-x_1)+(y_2-y_1)}$

Substituting the values we have:

$$\begin{gathered} PQ=\sqrt{{\left(\frac{-7}{6}-\frac{17}{6}\right)}^2+{\left(\frac{-1}{3}-\frac{8}{3}\right)}^2} \\ \Rightarrow PQ=\sqrt{{\left(\frac{-24}{6}\right)}^2+{\left(\frac{-9}{3}\right)}^2} \\ \Rightarrow PQ=\sqrt{16+9} \\ \Rightarrow PQ=\sqrt{25} \\ \Rightarrow PQ=5 \end{gathered}$$

Hence, the length of a line segment $PQ$ is equal to $5$