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

$\therefore PA:AQ=1:2$

Thus A divides PQ internally in the ratio 1 : 2

$\therefore A=(\frac{1\times 10+2\times 4}{1+2},\frac{1\times (-16)+2\times 2}{1+2},\frac{1\times 6+2\times (-6)}{1+2})$

$\equiv (\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3})$

$\equiv (\frac{18}{3},\frac{-12}{3},\frac{-6}{3})$

$\equiv (6,-4,-2)$

$\Rightarrow A\equiv (6,-4,-2)$

AB = BQ

Therefore, B is the mid-point of AQ.

$\therefore B\equiv (\frac{6+10}{2},\frac{-4-16}{2},\frac{-2+6}{2})$

$\equiv (\frac{16}{2},\frac{-20}{2},\frac{4}{2})$

$\equiv$ (8, -10, 2)