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

(a) Write down the coordinates of A and D.

(b) Find the coordinates of the centroid of ∆ABC.

(c) If D divides AC in the ratio k : 1, find the value of k.

(d) Find the equation of the line BD.

(a) From graph,

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

(b) By formula,

Co-ordinates of centroid = $\Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

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

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

(c) By section-formula,

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

Given,

D divides AC in the ratio k : 1.

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

Hence, k = 3.

(d) By two point form,

Equation of line :

y - y₁ = $\dfrac{y$2 - y1}{x2 - x1}(x - x_1)

Equation of BD :

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

⇒ y + 4 = $\dfrac{4}{-7}(x - 4)$

⇒ -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.