Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Let C, D and E divide the line segment AB into four equal parts.

By section-formula,

(x, y) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Let co-ordinates of C be (a, b) and from figure we see that C divides AB in the ratio 1 : 3.

Substituting values we get :

$\Rightarrow (a, b) = \Big(\dfrac{1 \times 2 + 3 \times -2}{1 + 3}, \dfrac{1 \times 8 + 3 \times 2}{1 + 3}\Big) \\[1em] = \Big(\dfrac{2 - 6}{4}, \dfrac{8 + 6}{4}\Big) \\[1em] = \Big(-\dfrac{4}{4}, \dfrac{14}{4}\Big) \\[1em] = \Big(-1, \dfrac{7}{2}\Big).$

From figure,

D is the mid-point of AB. Let co-ordinates of D be (g, h).

By formula,

Mid-point = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values we get :

$\Rightarrow (g, h) = \Big(\dfrac{-2 + 2}{2}, \dfrac{2 + 8}{2}\Big) \\[1em] = \Big(\dfrac{0}{2}, \dfrac{10}{2}\Big) \\[1em] = (0, 5).$

Let co-ordinates of E be (p, q) and from figure we see that E divides AB in the ratio 3 : 1.

Substituting values we get :

$\Rightarrow (p, q) = \Big(\dfrac{3 \times 2 + 1 \times -2}{1 + 3}, \dfrac{3 \times 8 + 1 \times 2}{1 + 3}\Big) \\[1em] = \Big(\dfrac{6 - 2}{4}, \dfrac{24 + 2}{4}\Big) \\[1em] = \Big(\dfrac{4}{4}, \dfrac{26}{4}\Big) \\[1em] = \Big(1, \dfrac{13}{2}\Big).$

Hence, co-ordinates of required points are $\Big(-1, \dfrac{7}{2}\Big), (0, 5), \Big(1, \dfrac{13}{2}\Big)$.