The vertices of a triangle are $A(10,4),B(-4,9)$ and $C(-2,-1)$. Find the equation of the altitude through $A$.

Given vertices of the triangle $△ABC$ are $A(10,4),B(-4,9),C(-2,-1).$

Perpendicular drawn from a vertex of a triangle to its opposite side is called Altitude.

Now, altitude from $A$ will be perpendicular to $BC.$

The formula for slope of a line with points $\left(x,y\right),\left(x\text{'},y\text{'}\right)$ is $\frac{y\text{'}-y}{x\text{'}-x}.$ 

$\Rightarrow$Slope of $BC=\frac{-1-9}{-2+4}=\frac{-10}{2}=-5.$

The multiplication of slopes of two perpendicular lines is $-1.$

$\Rightarrow$Slope of altitude $AD=\frac{-1}{slope of BC}=-\frac{1}{-5}=\frac{1}{5}$

The equation of line passing through the point $\left(x\text{'},y\text{'}\right)$ with slope $m$ is $y-y\text{'}=m\left(x-x\text{'}\right)$.

Hence, equation of altitude $AD$ which passes through $(10,4)$ and having slope $\frac{1}{5}$ is

$$\begin{gathered} \Rightarrow y-4=\frac{1}{5}\left(x-10\right) \\ \Rightarrow 5y-4=x-10 \\ \Rightarrow x-5y-6=0 \end{gathered}$$

Thus, the equation of altitude $AD$ is $x-5y-6=0.$

Therefore, the correct answer is option $1.$