ABC is a triangle whose vertices are A(1, –1), B(0, 4) and C(–6, 4). D is the mid-point of BC. Find the :

(i) co-ordinates of D

(ii) equation of the median AD

(i) By formula,

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

Given,

D is the mid-point of BC.

∴ Co-ordinates of D

$= \Big(\dfrac{0 + (-6)}{2}, \dfrac{4 + 4}{2}\Big) \\[1em] = \Big(\dfrac{-6}{2}, \dfrac{8}{2}\Big) \\[1em] = (-3, 4).$

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

(ii) Slope = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{4 - (-1)}{(-3) - 1} = -\dfrac{5}{4}

Equation of a line :

y - y₁ = m(x - x₁)

Substituting values we get :

Equation of AD :

⇒ y - (-1) = $-\dfrac{5}{4}$ (x - 1)

⇒ -4(y + 1) = 5(x - 1)

⇒ -4y - 4 = 5x - 5

⇒ 5x + 4y = -4 + 5

⇒ 5x + 4y - 1 = 0.

Hence, equation of median AD is 5x + 4y - 1 = 0.