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

(i) the equation of median BD,

(ii) the equation of altitude AE.

(i) D is the mid-point of AC [∵ BD is median]

By mid-point formula,

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

By two-point formula,

Equation of line : $y - y$1 = \dfrac{y2 - y1}{x2 - x1}(x - x1)

∴ Equation of line passing through (4, 4) and $\Big(-1, -\dfrac{1}{2}\Big)$.

$\Rightarrow y - 4 = \dfrac{-\dfrac{1}{2} - 4}{-1 - 4} \times (x - 4) \\[1em] \Rightarrow y - 4 = \dfrac{\dfrac{-1 - 8}{2}}{-5} \times (x - 4) \\[1em] \Rightarrow y - 4 = \dfrac{-\dfrac{9}{2}}{-5} \times (x - 4) \\[1em] \Rightarrow y - 4 = \dfrac{9}{10}(x - 4) \\[1em] \Rightarrow 10(y - 4) = 9(x - 4) \\[1em] \Rightarrow 10y - 40 = 9x - 36 \\[1em] \Rightarrow 9x - 10y - 36 + 40 = 0 \\[1em] \Rightarrow 9x - 10y + 4 = 0 \\[1em] \Rightarrow 9x + 4 = 10y.$

Hence, equation of median BD is 10y = 9x + 4.

(ii) By formula,

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

Slope of BC = $\dfrac{-2 - 4}{1 - 4} = \dfrac{-6}{-3}$ = 2.

We know that,

Product of slope of perpendicular lines = -1.

⇒ Slope of BC × Slope of AE = -1

⇒ 2 × Slope of AE = -1

⇒ Slope of AE = -$\dfrac{1}{2}$.

By point slope form, equation of AE :

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 1 = -\dfrac{1}{2}[x - (-3)] \\[1em] \Rightarrow 2(y - 1) = -[x + 3] \\[1em] \Rightarrow 2y - 2 = -x - 3 \\[1em] \Rightarrow 2y + x = -3 + 2 \\[1em] \Rightarrow x + 2y = -1 \\[1em] \Rightarrow x + 2y + 1 = 0.

Hence, equation of altitude AE is x + 2y + 1 = 0.