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) coordinates of point D.

(b) coordinates of point A.

(c) equation of a line parallel to line BC, through M.

(a) Given,

BD : DC = 2 : 1.

Let coordinates of D be (x, y)

By section-formula,

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

Substituting values, we get :

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

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

(b) By mid-point formula,

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

Given,

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

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

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

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

(c) By formula,

Slope of line = $\dfrac{y$2 - y1}{x2 - x1}

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

We know that,

Slope of parallel lines are equal.

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

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 = $-\dfrac{3}{4}(x - 6)$

⇒ y = $-\dfrac{3}{4}(x - 6)$

⇒ 4y = -3(x - 6)

⇒ 4y = -3x + 18

⇒ 3x + 4y = 18.

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