Question

Points P, Q and R, in that order, divide a line segment joining A(1, 6) and B(5, −2) in four equal parts. Find the coordinates of P, Q and R.

Answer 1

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ \Rightarrow x=\frac{\left(1\times 5+3\times 1\right)}{1+3},y=\frac{\left(1\times \left(-2\right)+3\times 6\right)}{1+3} \\ \Rightarrow x=\frac{5+3}{4},y=\frac{-2+18}{4} \\ \Rightarrow x=\frac{8}{4},y=\frac{16}{4} \\ \Rightarrow x=2\mathrm{and}y=4 \end{gathered}$$
Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):
$$\begin{gathered} x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2} \\ \Rightarrow x=\frac{1+5}{2},y=\frac{6+\left(-2\right)}{2} \\ \Rightarrow x=\frac{6}{2},y=\frac{4}{2} \\ \Rightarrow x=3,y=2 \end{gathered}$$
Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:
$$\begin{gathered} x=\frac{\left(m{x}_2+n{x}_1\right)}{\left(m+n\right)},y=\frac{\left(m{y}_2+n{y}_1\right)}{\left(m+n\right)} \\ \Rightarrow x=\frac{\left(3\times 5+1\times 1\right)}{3+1},y=\frac{\left(3\times \left(-2\right)+1\times 6\right)}{3+1} \\ \Rightarrow x=\frac{15+1}{4},y=\frac{-6+6}{4} \\ \Rightarrow x=\frac{16}{4},y=\frac{0}{4} \\ \Rightarrow x=4\mathrm{and}y=0 \end{gathered}$$
Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.