The vertices of a △ABC are A(3, 8), B(-1, 2) and C(6, -6). Find:

(i) slope of BC.

(ii) equation of a line perpendicular to BC and passing through A.

(i) Let the slope of BC be m₁. Slope of BC is given by,

$\text{m}$1 = \dfrac{y2 - y1}{x2 - x_1} \\[1em] = \dfrac{-6 - 2}{6 - (-1)} \\[1em] = -\dfrac{8}{7}.

Hence, the slope of BC is $-\dfrac{8}{7}.$

(ii) Let slope of line perpendicular to BC be m₂.

So, m₁ × m₂ = -1.

$\Rightarrow -\dfrac{8}{7} \times m$2 = -1 \\[1em] \Rightarrow m2 = \dfrac{7}{8}.

Equation of the line having the slope = $\dfrac{7}{8}$ and passing through A(3, 8) can be given by point-slope formula i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 8 = \dfrac{7}{8}(x - 3) \\[1em] \Rightarrow 8(y - 8) = 7(x - 3) \\[1em] \Rightarrow 8y - 64 = 7x - 21 \\[1em] \Rightarrow 7x - 8y - 21 + 64 = 0 \\[1em] \Rightarrow 7x - 8y + 43 = 0.

Hence, the equation of the required line is 7x - 8y + 43 = 0.