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

Solution:

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:

$x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)}, y=\frac{\left(m y_{2}+n y_{1}\right)}{(m+n)}$

$\Rightarrow x=\frac{(1 \times 5+3 \times 1)}{1+3}, y=\frac{(1 \times(-2)+3 \times 6)}{1+3}$

$\Rightarrow x=\frac{5+3}{4}, y=\frac{-2+18}{4}$

$\Rightarrow x=\frac{8}{4}, y=\frac{16}{4}$

$\Rightarrow x=2$ and $y=4$

Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):

$x=\frac{x_{1}+x_{2}}{2}, y=\frac{y_{1}+y_{2}}{2}$

$\Rightarrow x=\frac{1+5}{2}, y=\frac{6+(-2)}{2}$

$\Rightarrow x=\frac{6}{2}, y=\frac{4}{2}$

$\Rightarrow x=3, y=2$

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:

$x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)}, y=\frac{\left(m y_{2}+n y_{1}\right)}{(m+n)}$

$\Rightarrow x=\frac{(3 \times 5+1 \times 1)}{3+1}, y=\frac{(3 \times(-2)+1 \times 6)}{3+1}$

$\Rightarrow x=\frac{15+1}{4}, y=\frac{-6+6}{4}$

$\Rightarrow x=\frac{16}{4}, y=\frac{0}{4}$

$\Rightarrow x=4$ and $y=0$

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.