A(–4, 2), B(6, 4) and C(2, –2) are the vertices of ΔABC. Find :

(i) the equation of median AD

(ii) the equation of altitude BM

(iii) the equation of right bisector of AB

(iv) the co-ordinates of centroid of ΔABC

(i) Slope of AD = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{1 - 2}{4 - (-4)} = \dfrac{-1}{8}

A median joins a vertex to the midpoint of the opposite side. D is the midpoint of BC.

$= \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{6 + 2}{2}, \dfrac{4 + (-2)}{2}\Big) \\[1em] = \Big(\dfrac{8}{2}, \dfrac{2}{2}\Big) = (4, 1).

By point-slope form,

Equation of a median AB, given by:

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = $\dfrac{-1}{8}$ [x - (-4)]

⇒ 8(y - 2) = -1(x + 4)

⇒ 8y - 16 = -x - 4

⇒ x + 8y - 12 = 0

Hence, the equation of a line AD x + 8y - 12 = 0.

(ii) Slope of AC = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{-2 - 2}{2 - (-4)} = \dfrac{-4}{6} = -\dfrac{2}{3}

We know that altitude BM is a perpendicular AC.

Let the slope of BM be m₁,

⇒ m_AC × m₂ = -1

⇒ $-\dfrac{2}{3}$ × m₁ = -1

⇒ m₁ = $\dfrac{3}{2}$

Equation of a line BM,

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = $\dfrac{3}{2}$ (x - 6)

⇒ 2(y - 4) = 3(x - 6)

⇒ 2y - 8 = 3x - 18

⇒ 3x - 2y - 10 = 0

Hence, the equation of a line BM 3x - 2y - 10 = 0.

(iii) Slope of AB = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{4 - 2}{6 - (-4)} = \dfrac{2}{10} = \dfrac{1}{5}

Right bisector of AB is perpendicular to AB

Let the slope of Right bisector of AB be m₂,

⇒ m_AB × m₂ = -1

⇒ $\dfrac{1}{5}$ × m₂ = -1

⇒ m₂ = -5

Coordinates of Midpoint of AB

$= \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{-4 + 6}{2}, \dfrac{2 + 4}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{6}{2}\Big) = (1, 3).

Equation of a line BM,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -5 (x - 1)

⇒ (y - 3) = -5x + 5

⇒ 5x + y - 8 = 0

Hence, the equation of right bisector of AB 5x + y - 8 = 0.

(iv) Centroid of triangle ABC = $\Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

Substitute values we get,

$= \Big(\dfrac{-4 + 6 + 2}{3}, \dfrac{2 + 4 + (-2)}{3}\Big) = \Big(\dfrac{4}{3}, \dfrac{4}{3}\Big)$

Hence, coordinates of centroid are $\Big(\dfrac{4}{3}, \dfrac{4}{3}\Big)$.