Question

Find the coordinates of the points which divide the line segment joining the points (-2,2) and (6,-6) in four equal parts.

Answer 1

$$\begin{gathered} \mathrm{Le}\text{t}\text{the}\mathrm{points}\mathrm{P},\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{divide}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{AB}\mathrm{in}\mathrm{four}\mathrm{equal}\mathrm{parts}. \\ \mathrm{Therefore},\mathrm{point}\mathrm{Q}\mathrm{divides}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{ratio}2:2=1:1,\mathrm{internally}. \\ \mathrm{Let}\mathrm{Q}\equiv \left(x_2,y_2\right)\mathrm{which}\mathrm{divides}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{ratio}1:1. \\ \mathrm{That}\mathrm{is},\mathrm{point}\mathrm{Q}\mathrm{bisects}\mathrm{seg}\mathrm{AB}. \\ \text{H}\mathrm{ere},\mathrm{A}\equiv \left(-2,2\right)\mathrm{and}\mathrm{B}\equiv \left(6,-6\right). \end{gathered}$$

$$\begin{gathered} \therefore x_2=\frac{-2+6}{2}=2 \\ \mathrm{and}{y}_2=\frac{2-6}{2}=-2 \\ \therefore Q\equiv \left(2,-2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly},\mathrm{point}\mathrm{P}\left(x_1,y_1\right)\mathrm{bisects}\mathrm{seg}\mathrm{AQ} \\ {x}_1=\frac{-2+2}{2}=0 \\ {y}_1=\frac{2-2}{2}=0 \\ \therefore \mathrm{P}\equiv \left(0,0\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly},\mathrm{point}\mathrm{R}\left(x_3,y_3\right)\mathrm{bisects}\mathrm{seg}\mathrm{QB} \\ {x}_1=\frac{2+6}{2}=4 \\ {y}_1=\frac{-2-6}{2}=-4 \\ \therefore \mathrm{R}\equiv \left(4,-4\right) \end{gathered}$$