Question

If A (–14, –10), B(6, –2) is given, find the coordinates of the points which divide segment AB into four equal parts.

Answer 1

Let $\mathrm{P}\left(x_1,y_1\right),\mathrm{Q}\left(x_2,y_2\right),\mathrm{R}\left(x_3,y_3\right)$ be the points which divide the line segment AB into 4 equal parts.
AP = PQ = QR = RB
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{RB}}=\frac{\mathrm{AP}}{\mathrm{AP}+\mathrm{AP}+\mathrm{AP}}=\frac{\mathrm{AP}}{3\mathrm{AP}}=\frac{1}{3}$
$$\begin{gathered} x_1=\left(\frac{6\times 1+3\times \left(-14\right)}{1+3}\right)=\frac{6-42}{4}=-9 \\ {y}_1=\left(\frac{1\times \left(-2\right)+3\times \left(-10\right)}{1+3}\right)=\frac{-2-30}{4}=-8 \\ \mathrm{P}\left(x_1,y_1\right)=\left(-9,-8\right) \end{gathered}$$
Now, $\frac{\mathrm{PQ}}{\mathrm{QB}}=\frac{\mathrm{PQ}}{\mathrm{QR}+\mathrm{RB}}=\frac{\mathrm{PQ}}{\mathrm{PQ}+\mathrm{PQ}}=\frac{1}{2}$
$$\begin{gathered} x_2=\left(\frac{1\times 6+2\times \left(-9\right)}{1+2}\right)=-4 \\ {y}_2=\left(\frac{1\times \left(-2\right)+2\times \left(-8\right)}{1+2}\right)=-6 \\ \mathrm{Q}\left(x_2,y_2\right)=\left(-4,-6\right) \end{gathered}$$
Now R divides QB into 2 equal parts so, using the midpoint formula we have
$$\begin{gathered} x_3=\frac{-4+6}{2}=1 \\ {y}_3=\frac{-6+\left(-2\right)}{2}=-4 \\ \mathrm{R}\left(x_3,y_3\right)=\left(1,-4\right) \end{gathered}$$
Thus, $\mathrm{P}\left(x_1,y_1\right),\mathrm{Q}\left(x_2,y_2\right),\mathrm{R}\left(x_3,y_3\right)$ are $\left(-9,-8\right),\left(-4,-6\right)\mathrm{and}\left(1,-4\right)$.