A(a, b), B(-4, 3) and C(8, -6) are the vertices of a △ ABC. Point D is on BC such that BD : DC is 2 : 1 and M(6, 0) is mid-point of AD. Find :

(a) Given,

BD : DC = 2 : 1.

Let coordinates of D be (x, y)

By section-formula,

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

Substituting values, we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{2 \times 8 + 1 \times (-4)}{2 + 1}, \frac{2 \times -6 + 1 \times 3}{2 + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\frac{16 - 4}{3}, \frac{-12 + 3}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\frac{12}{3}, \frac{-9}{3}\Big) \\[1em] \Rightarrow (x, y) = (4, -3).\end{aligned}$$

Hence, coordinates of D = (4, -3).

(b) By mid-point formula,

Mid-point = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Given,

M(6, 0) is the mid-point of AD.

$$\begin{aligned}\therefore (6, 0) = \Big(\frac{a + 4}{2}, \frac{b + (-3)}{2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\frac{a + 4}{2}, \frac{b - 3}{2}\Big) \\[1em] \Rightarrow \frac{a + 4}{2} = 6 \text{ and } \frac{b - 3}{2} = 0 \\[1em] \Rightarrow a + 4 = 12 \text{ and } b - 3 = 0 \\[1em] \Rightarrow a = 12 - 4 = 8 \text{ and } b = 3.\end{aligned}$$

A = (a, b) = (8, 3).

Hence, coordinates of A = (8, 3).

(c) By formula,

Slope of line = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope of line BC = $\frac{-6 - 3}{8 - (-4)} = \frac{-9}{12} = -\frac{3}{4}$.

We know that,

Slope of parallel lines are equal.

Slope of line parallel to BC = $-\frac{3}{4}$.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Equation of line parallel to BC and passing through M is :

⇒ y - 0 = $-\frac{3}{4}(x - 6)$

⇒ y = $-\frac{3}{4}(x - 6)$

⇒ 4y = -3(x - 6)

⇒ 4y = -3x + 18

⇒ 3x + 4y = 18.

Hence, equation of the required line is 3x + 4y = 18.