Question

If A (20, 10), B(0, 20) are given, find the coordinates of the points which divide segment AB into five congruent parts.

Answer 1

Let the points $\mathrm{P}\left(x_1,y_1\right),\mathrm{Q}\left(x_2,y_2\right),\mathrm{R}\left(x_3,y_3\right)\mathrm{and}\mathrm{S}\left(x_4,y_4\right)$ be the points which divide the line segment AB into 5 equal parts.
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{RS}}=\frac{\mathrm{AP}}{4\mathrm{AP}}=\frac{1}{4}$
$$\begin{gathered} x_1=\left(\frac{1\times 0+4\times 20}{1+4}\right)=16 \\ {y}_1=\left(\frac{1\times 20+4\times 10}{1+4}\right)=12 \\ \mathrm{P}\left(x_1,y_1\right)=\left(16,12\right) \end{gathered}$$
$\frac{\mathrm{PQ}}{\mathrm{QB}}=\frac{\mathrm{PQ}}{\mathrm{QR}+\mathrm{RS}+\mathrm{SB}}=\frac{\mathrm{PQ}}{\mathrm{PQ}+\mathrm{PQ}+\mathrm{PQ}}=\frac{\mathrm{PQ}}{3\mathrm{PQ}}=\frac{1}{3}$
$$\begin{gathered} x_2=\left(\frac{1\times 0+3\times 16}{1+3}\right)=12 \\ {y}_2=\left(\frac{1\times 20+3\times 12}{1+3}\right)=14 \\ \mathrm{Q}\left(x_2,y_2\right)=\left(12,14\right) \end{gathered}$$
$\frac{\mathrm{QR}}{\mathrm{RB}}=\frac{\mathrm{QR}}{\mathrm{RS}+\mathrm{SB}}=\frac{\mathrm{QR}}{\mathrm{QR}+\mathrm{QR}}=\frac{\mathrm{QR}}{2\mathrm{QR}}=\frac{1}{2}$
$$\begin{gathered} x_3=\left(\frac{1\times 0+2\times 12}{1+2}\right)=8 \\ {y}_3=\left(\frac{1\times 20+2\times 14}{1+3}\right)=16 \\ \mathrm{R}\left(x_3,y_3\right)=\left(8,16\right) \end{gathered}$$
S is the midpoint of RB so, using the midpoint formula
$$\begin{gathered} x_4=\frac{8+0}{2}=4 \\ {y}_4=\frac{16+20}{2}=18 \\ \mathrm{S}\left(x_4,y_4\right)=\left(4,18\right) \end{gathered}$$
So, the points
$$\begin{gathered} \mathrm{P}\left(x_1,y_1\right)=\left(16,12\right) \\ \mathrm{Q}\left(x_2,y_2\right)=\left(12,14\right) \\ \mathrm{R}\left(x_3,y_3\right)=\left(8,16\right) \\ \mathrm{S}\left(x_4,y_4\right)=\left(4,18\right) \end{gathered}$$