Question

Let \[A\left( {1,0} \right)\], \[B\left( {6,2} \right)\] and \[C\left( {\dfrac{3}{2},6} \right)\] be the vertices of a triangle \[ABC\]. If \[P\] is a point inside the triangle \[ABC\] such that the triangle \[APC\], \[APB\] and \[BPC\] have equal areas. Then what is the length of the line the segment \[PQ\], where \[Q\] is the point \[\left( {\dfrac{{ - 7}}{6},\dfrac{{ - 1}}{3}} \right)\]?

Answer 1

Hint: Centroid divides the triangle into 3 sub-triangles of equal area. Using the coordinates of a triangle, find the centroid of a triangle. Then use the distance formula to get the required answer.

Formula used:
The centroid of a triangle with vertices \[\left( {{x_1},{y_1}} \right)\], \[\left( {{x_2},{y_2}} \right)\] and \[\left( {{x_3},{y_3}} \right)\] is:\[\left( {x,y} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\]
Distance formula:
The distance between the two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is:
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]

Complete step by step solution:
Given:
The vertices of the triangles are \[A\left( {1,0} \right)\], \[B\left( {6,2} \right)\] and \[C\left( {\dfrac{3}{2},6} \right)\].
The point \[P\] divides the triangle \[ABC\] such that \[APC\], \[APB\] and \[BPC\] have equal areas.
We know that the centroid of a triangle divides the triangle into three sub-triangles with equal area.
So, \[P\] is the centroid of the triangle \[ABC\].
Apply the formula of the centroid of the triangle.
\[P\left( {x,y} \right) = \left( {\dfrac{{1 + 6 + \dfrac{3}{2}}}{3},\dfrac{{0 + 2 + 6}}{3}} \right)\]
\[ \Rightarrow \]\[P\left( {x,y} \right) = \left( {\dfrac{{7 + \dfrac{3}{2}}}{3},\dfrac{8}{3}} \right)\]
Simplify the coordinates.
\[P\left( {x,y} \right) = \left( {\dfrac{{\dfrac{{17}}{2}}}{3},\dfrac{8}{3}} \right)\]
\[ \Rightarrow \]\[P\left( {x,y} \right) = \left( {\dfrac{{17}}{6},\dfrac{8}{3}} \right)\]

Image: The diagram of triangle \[ABC\] with centroid \[P\] and a point \[Q\]
Now apply the distance formula to calculate the length of the line segment \[PQ\].
\[PQ = \sqrt {{{\left( {\dfrac{{17}}{6} - \left( {\dfrac{{ - 7}}{6}} \right)} \right)}^2} + {{\left( {\dfrac{8}{3} - \left( {\dfrac{{ - 1}}{3}} \right)} \right)}^2}} \]
\[ \Rightarrow \]\[PQ = \sqrt {{{\left( {\dfrac{{17}}{6} + \dfrac{7}{6}} \right)}^2} + {{\left( {\dfrac{8}{3} + \dfrac{1}{3}} \right)}^2}} \]
Solve the above equation.
\[PQ = \sqrt {{{\left( {\dfrac{{24}}{6}} \right)}^2} + {{\left( {\dfrac{9}{3}} \right)}^2}} \]
\[ \Rightarrow \]\[PQ = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2}} \]
\[ \Rightarrow \]\[PQ = \sqrt {16 + 9} \]
\[ \Rightarrow \]\[PQ = \sqrt {25} \]
\[ \Rightarrow \]\[PQ = \pm 5\]
The distance between any points is positive.
Thus, \[PQ = 5\].
Hence, the length of the line segment \[PQ\] is 5 units.

Note: Centroid of a triangle is the point of intersection of the all 3 medians of a triangle and positioned inside the triangle.
At centroid, each median in a triangle is divided in the ratio \[2:1\].