Question

Find the coordinates of the points of trisection of the line segment AB, whose end points are A(2, 1) and B(5, −8).

Answer 1

Let P and Q be the points of trisection of AB.
Then P divides AB in the ratio 1:2.
So the coordinates of P are
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ x=\frac{\left(1\times 5+2\times 2\right)}{1+2},y=\frac{\left\{1\times \left(-8\right)+2\times 1\right\}}{1+2} \\ x=\frac{5+4}{3},y=\frac{-8+2}{3} \\ x=\frac{9}{3},y=-\frac{6}{3} \\ \mathrm{So},x=3\mathrm{and}y=-2 \end{gathered}$$
Therefore, the coordinates of P are (3, −2).
Also, Q divides AB in the ratio 2:1.
So, the coordinates of Q are
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ x=\frac{\left(2\times 5+1\times 2\right)}{2+1},y=\frac{\left\{2\times \left(-8\right)+1\times 1\right\}}{2+1} \\ x=\frac{10+2}{3},y=\frac{-16+1}{3} \\ x=\frac{12}{3},y=-\frac{15}{3} \\ \mathrm{So},x=4\mathrm{and}y=-5 \end{gathered}$$
Therefore, the coordinates of Q are (4, −5).
Hence, the points are (3, −2) and (4, −5).