Question

Find the coordinates of the points which tisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6).

Answer 1

Let P$\equiv$(4, 2, $-$6) and Q$\equiv$(10, $-$16, 6)
Let A and B be the point of trisection.
Then, we have:
PA = AB = BQ
∴ PA:AQ = 1:2
Thus, A divides PQ internally in the ratio 1:2.
∴ A$\equiv$$\left(\frac{1\times 10+2\times 4}{1+2},\frac{1\times \left(-16\right)+2\times 2}{1+2},\frac{1\times 6+2\times \left(-6\right)}{1+2}\right)$
$$\begin{gathered} \equiv \left(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3}\right) \\ \equiv \left(\frac{18}{3},\frac{-12}{3},\frac{-6}{3}\right) \\ \equiv \left(6,-4,-2\right) \end{gathered}$$
$\Rightarrow$ A$\equiv$(6, $-$4, $-$2)
AB = BQ



Therefore, B is the mid-point of AQ.
∴ B$\equiv$$\left(\frac{6+10}{2},\frac{-4-16}{2},\frac{-2+6}{2}\right)$
$$\begin{gathered} \equiv \left(\frac{16}{2},\frac{-20}{2},\frac{4}{2}\right) \\ \equiv \left(8,-10,2\right) \end{gathered}$$
$\Rightarrow$ B$\equiv$(8, $-$10, 2)