Question

Find the coordinates of the points which trisect the line segment joining the points $P\left( 4,2,-6 \right)$ and $Q\left( 10,-16,6 \right)$

Answer 1

To obtain the coordinates of the points which trisect the line segment joining the given point we will use section formula. Firstly we will let two points in the line to be $A$ and $B$ such that each segment is of equal length. Then we will find the coordinate of the point we let by using section formula and get our desired answer.

Complete step by step answer:
The line segment joining the two points is:
$P=\left( 4,2,-6 \right)$……$\left( 1 \right)$
$Q=\left( 10,-16,6 \right)$……$\left( 2 \right)$

Let the two points which trisect the given line be $A$ and $B$ such that:
$PA=AB=BQ$

So $A$ divides the line $PQ$ in $1:2$ and $B$ divide the line in $2:1$
The section formula is given as:
\left( x,y,z \right)=\left\\{ \left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \right),\left( \dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right),\left( \dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right) \right\\}……….$\left( 3 \right)$
Where $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ are the coordinates of the two points given and $m,n$ is the ratio in which the point divide the given line.
So on comparing with equation (1) and (2) we get,
$\begin{aligned} & \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 4,2,-6 \right) \\\ & \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 10,-16,6 \right) \\\ \end{aligned}$
For point $A$ we have the ratio as $1:2$so
$\begin{aligned} & m=1 \\\ & n=2 \\\ \end{aligned}$
$\begin{aligned} & \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 4,2,-6 \right) \\\ & \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 10,-16,6 \right) \\\ \end{aligned}$
Substitute above values in equation (3) we get,
\begin{aligned} & A\left( x,y,z \right)=\left\\{ \left( \dfrac{1\times 10+2\times 4}{1+2} \right),\left( \dfrac{1\times -16+2\times 2}{1+2} \right),\left( \dfrac{1\times 6+2\times -6}{1+2} \right) \right\\} \\\ & A\left( x,y,z \right)=\left\\{ \left( \dfrac{10+8}{3} \right),\left( \dfrac{-16+4}{3} \right),\left( \dfrac{6-12}{3} \right) \right\\} \\\ & A\left( x,y,z \right)=\left( 6,-4,-2 \right) \\\ \end{aligned}
For point $B$ we have the ratio as $2:1$so
$\begin{aligned} & m=2 \\\ & n=1 \\\ \end{aligned}$
$\begin{aligned} & \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 4,2,-6 \right) \\\ & \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( 10,-16,6 \right) \\\ \end{aligned}$
Substitute above values in equation (3) we get,
\begin{aligned} & B\left( x,y,z \right)=\left\\{ \left( \dfrac{2\times 10+1\times 4}{2+1} \right),\left( \dfrac{2\times -16+1\times 2}{2+1} \right),\left( \dfrac{2\times 6+1\times -6}{2+1} \right) \right\\} \\\ & B\left( x,y,z \right)=\left\\{ \left( \dfrac{20+4}{3} \right),\left( \dfrac{-32+2}{3} \right),\left( \dfrac{12-6}{3} \right) \right\\} \\\ & B\left( x,y,z \right)=\left( 8,-10,2 \right) \\\ \end{aligned}
So we got the points as $A\left( 6,-4,-2 \right)$ and $B\left( 8,-10,2 \right)$
Hence coordinates of the points which trisect the line segment joining the points $P\left( 4,2,-6 \right)$ and $Q\left( 10,-16,6 \right)$ is $\left( 6,-4,-2 \right)$ and $\left( 8,-10,2 \right)$

Note: Section formula is used in coordinate geometry topic to find out the ratio in which a line segment is divided by any point internally or externally. Section formula can be used both in 3-D as well as 2-D planes. We should know the coordinate of the points which are forming the line segment to find out the ratio in which the points divide the line.