Precalculus (6th Edition) Blitzer

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 $4x+y+3=0$.
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 $4x+y+3=0$.