In the given diagram, ABC is a triangle, where B(4, -4) and C(-4, -2). D is a point on AC.

(a) From graph,

Co-ordinates of A = (0, 6) and D = (-3, 0)

(b) By formula,

Co-ordinates of centroid = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

$$\begin{aligned}= \Big(\frac{0 + 4 + (-4)}{3}, \frac{6 + (-4) + (-2)}{3}\Big) \\[1em] = \Big(\frac{0}{3}, \frac{0}{3}\Big) \\[1em] = (0, 0).\end{aligned}$$

Hence, co-ordinates of centroid of ∆ABC = (0, 0).

(c) By section-formula,

(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Given,

D divides AC in the ratio k : 1.

$$\begin{aligned}\therefore (-3, 0) = \Big(\frac{k \times -4 + 1 \times 0}{k + 1}, \frac{k \times -2 + 1 \times 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\frac{-4k + 0}{k + 1}, \frac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\frac{-4k}{k + 1}, \frac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow -3 = -\frac{4k}{k + 1} \text{ and } 0 = \frac{-2k + 6}{k + 1} \\[1em] \Rightarrow -3(k + 1) = -4k \text{ and } 0 = -2k + 6 \\[1em] \Rightarrow 3(k + 1) = 4k \text{ and } 2k = 6 \\[1em] \Rightarrow 3k + 3 = 4k \text{ and } k = \frac{6}{2} \\[1em] \Rightarrow 4k - 3k = 3 \text{ and } k = 3 \\[1em] \Rightarrow k = 3.\end{aligned}$$

Hence, k = 3.

(d) By two point form,

Equation of line :

y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of BD :

⇒ y - (-4) = $\frac{0 - (-4)}{-3 - 4}(x - 4)$

⇒ y + 4 = $\frac{4}{-7}(x - 4)$

⇒ -7(y + 4) = 4(x - 4)

⇒ -7y - 28 = 4x - 16

⇒ 4x + 7y - 16 + 28 = 0

⇒ 4x + 7y + 12 = 0.

Hence, equation of BD is 4x + 7y + 12 = 0.