A(5, 4), B(-3, -2) and C(1, -8) are the vertices of a triangle ABC. Find :

(i) the slope of the altitude of AB,

(ii) the slope of the median AD and

(iii) the slope of the line parallel to AC.

(i) By formula,

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

$\text{Slope of AB }= \dfrac{-2 - 4}{-3 - 5} \\[1em] = \dfrac{-6}{-8} \\[1em] = \dfrac{3}{4}.$

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of altitude = -1

⇒ $\dfrac{3}{4}$ x Slope of altitude = -1

⇒ Slope of altitude = $-\dfrac{4}{3}$

Hence, slope of the altitude of AB = $-\dfrac{4}{3}$.

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

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

$\text{Slope of AD }= \dfrac{-5 - 4}{-1 - 5} \\[1em] = \dfrac{-9}{-6} \\[1em] = \dfrac{3}{2}.$

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

(iii) $\text{Slope of AC }= \dfrac{-8 - 4}{1 - 5} \\[1em] = \dfrac{-12}{-4} \\[1em] = 3.$

Since, slope of parallel lines are equal.

Hence, slope of line parallel to AC = 3.