#### Answer

The point-slope form of the line passing through $\left( -2,5 \right)$ and perpendicular to the line $x-4y-8$ is
$y-5=-4\left( x+2 \right)$ and the general form is $4x+y+3=0$.

#### Work Step by Step

Consider the equation, $x-4y+8=0$
The slope-intercept form of the equation is $x-4y+8=0$.
Collect all the x and y variables on either side of the equation.
$\begin{align}
& x-4y+8=0 \\
& -4y=-x-8 \\
& 4y=x+8
\end{align}$
Now isolate the y term for that divides both sides of the equation by $4$.
$y=\frac{1}{4}x+2$
Hence, the slope-intercept form of the equation is $y=\frac{1}{4}x+2$.
The provided line has slope $\frac{1}{4}$.
The product of the slope of the perpendicular line is $-1$.
${{m}_{1}}\cdot {{m}_{2}}=-1$
The line perpendicular to this line has a slope that is the negative reciprocal of $\frac{1}{4}$.
Thus, the slope of any perpendicular line is $-4$
The line passes through the point $\left( -2,5 \right)$
So, we have $x=-2\text{ and }y=5$.
The equation of the line in point-slope form is
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Therefore,
$\begin{align}
& y-\left( 5 \right)=m\left( x-\left( -2 \right) \right) \\
& y-5=-4\left( x+2 \right) \\
\end{align}$
For the general form, apply the distributive property as:
$y-5=-4x-8$
Simplify as:
$4x+y+3=0$
Hence, the point-slope form of the line passing through $\left( -2,5 \right)$ and perpendicular to the line $x-4y-8$ is $y-5=-4\left( x+2 \right)$ and the general form is $4x+y+3=0$.