Question

Find the coordinates of the points that divide the join of A(−4, 0) and B(0, 6) in three equal parts.

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
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ \Rightarrow x=\frac{\left\{1\times 0+2\times \left(-4\right)\right\}}{1+2},y=\frac{\left(1\times 6+2\times 0\right)}{1+2} \\ \Rightarrow x=-\frac{8}{3},y=\frac{6}{3} \\ \Rightarrow x=-\frac{8}{3},y=2 \\ \mathrm{Therefore},\text{coordinates of}P\text{are}\left(-\frac{8}{3},2\right)\text{.} \end{gathered}$$
Also Q divides AB in the ratio 2:1
So the coordinates of Q are
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ x=\frac{\left\{2\times 0+1\times \left(-4\right)\right\}}{2+1},y=\frac{\left(2\times 6+1\times 0\right)}{2+1} \\ x=-\frac{4}{3},y=\frac{12}{3} \\ x=-\frac{4}{3},y=4 \\ \mathrm{Therefore},\text{coordinates of}Q\text{are}\left(-\frac{4}{3},4\right). \end{gathered}$$
Hence, the points are $\left(-\frac{8}{3},2\right)\mathrm{and}\left(-\frac{4}{3},4\right)$.