The vertices of a triangle are A(10, 4), B(4, -9) and C(-2, -1). Find the equation of the altitude through A.
[The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.]

Given, vertices of a triangle are A(10, 4), B(4, -9) and C(-2, -1).

Now,

Slope of line BC (m₁),

$\text{m}$1 = \dfrac{y2 - y1}{x2 - x_1} \\[1em] = \dfrac{-1 - (-9)}{-2 - 4} \\[1em] = -\dfrac{8}{6} \\[1em] = -\dfrac{4}{3}.

Let the slope of the altitude from A(10, 4) to BC be m₂.

Then, m₁ × m₂ = -1.

$\Rightarrow -\dfrac{4}{3} \times m$2 = -1 \\[1em] \Rightarrow m2 = \dfrac{3}{4}.

Equation of the line having the slope = $\dfrac{3}{4}$ and passing through A(10, 4) can be given by point-slope formula i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 4 = \dfrac{3}{4}(x - 10) \\[1em] \Rightarrow 4(y - 4) = 3(x - 10) \\[1em] \Rightarrow 4y - 16 = 3x - 30 \\[1em] \Rightarrow 3x - 4y + 16 - 30 = 0 \\[1em] \Rightarrow 3x - 4y - 14 = 0.

Hence, the equation of the required line is 3x - 4y - 14 = 0.