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 :

(a) By formula,

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

Given,

D is the mid-point of BC.

$$\begin{aligned}\therefore \text{Co-ordinates of D} = \Big(\frac{0 + (-6)}{2}, \frac{4 + 4}{2}\Big) \\[1em] = \Big(\frac{-6}{2}, \frac{8}{2}\Big) \\[1em] = (-3, 4).\end{aligned}$$

Hence, co-ordinates of D = (-3, 4).

(b) By two-point form :

Equation of a line :

y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Substituting values we get :

Equation of AD :

⇒ y - (-1) = $\frac{4 - (-1)}{(-3) - 1}(x - 1)$

⇒ y + 1 = $\frac{5}{-4}(x - 1)$

⇒ -4(y + 1) = 5(x - 1)

⇒ -4y - 4 = 5x - 5

⇒ 5x + 4y = -4 + 5

⇒ 5x + 4y = 1.

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