In the given figure, if the line segment AB is intercepted by the y-axis and x-axis at C and D, respectively, such that AC : AD = 1 : 4 and D is the midpoint of CB. Find the coordinates of D, C and B.

Let coordinates of C be (0, b) and D be (a, 0).

Given,

AC : AD = 1 : 4

Let AC = x and AD = 4x.

From figure,

⇒ AD = AC + CD

⇒ 4x = x + CD

⇒ CD = 4x - x = 3x.

AC : CD = 1 : 3.

By section formula,

$\Rightarrow (x, y) = \Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big) \\[1em] \Rightarrow (0, b) = \Big(\dfrac{1 \times a + 3 \times -2}{1 + 3}, \dfrac{1 \times 0 + 3 \times 6}{1 + 3}\Big) \\[1em] \Rightarrow (0, b) = \Big(\dfrac{a - 6}{4}, \dfrac{0 + 18}{4}\Big) \\[1em] \Rightarrow (0, b) = \Big(\dfrac{a - 6}{4}, \dfrac{18}{4}\Big) \\[1em] \Rightarrow \dfrac{a - 6}{4} = 0 \text{ and } b = \dfrac{18}{4} \\[1em] \Rightarrow a - 6 = 0 \text{ and } b = \dfrac{9}{2} \\[1em] \Rightarrow a = 6 \text{ and } b = \dfrac{9}{2}.

C = (0 , b) = $\Big(0, \dfrac{9}{2}\Big)$ and D = (6, 0).

Given, D is the mid-point of CB.

$\therefore (6, 0) = \Big(\dfrac{0 + p}{2}, \dfrac{\dfrac{9}{2} + q}{2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\dfrac{p}{2}, \dfrac{9 + 2q}{2 \times 2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\dfrac{p}{2}, \dfrac{9 + 2q}{4}\Big) \\[1em] \Rightarrow \dfrac{p}{2} = 6 \text{ and } \dfrac{9 + 2q}{4} = 0 \\[1em] \Rightarrow p = 12 \text{ and } 9 + 2q = 0 \\[1em] \Rightarrow p = 12 \text{ and } 2q = -9 \\[1em] \Rightarrow p = 12 \text{ and } q = -\dfrac{9}{2} \\[1em] \Rightarrow B = (p, q) = \Big(12, -\dfrac{9}{2}\Big).$

Hence, coordinates of B = $\Big(12, -\dfrac{9}{2}\Big), C = \Big(0, \dfrac{9}{2}\Big)$ and D = (6, 0).