Given that A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a ΔABC. Find:

(i) the slope of median AD

(ii) the slope of altitude BM

(i) Since, AD is median. So, D is the mid-point of BC.

By using formula,

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

Substitute values we get,

D = $\Big(\dfrac{-3 + 1}{2}, \dfrac{-2 + (-8)}{2}\Big) = \Big(\dfrac{-2}{2}, \dfrac{-10}{2}\Big) = (-1, -5)$

By using slope formula,

m = $\dfrac{y$2 - y1}{x2 - x1}

Slope of AD = $\dfrac{-5 - 4}{-1 - 5} = \dfrac{-9}{-6} = \dfrac{3}{2}$

Hence, slope of the median AD = $\dfrac{3}{2}$.

(ii) The altitude BM is perpendicular to the side AC. Therefore, the product of their slopes is -1.

Slope of AC = $\dfrac{-8 - 4}{1 - 5} = \dfrac{-12}{-4} = 3$

m_BM × 3 = -1

m_BM = $\dfrac{-1}{3}$

Hence, slope of the BM = $-\dfrac{1}{3}$.