Chapter 13.7, Problem 28WE

To determine

To find : an equation of line described

Answer to Problem 28WE

The equation is $4x-3y-32=0$

Explanation of Solution

Given information :

Line through $\left(8,0\right)$ and perpendicular to the line $3x+4y=12$

Formula used :

  $y-y_1=m\left(x-x_1\right)$

Calculation :

The line

  $\begin{array}{l}3x+4y=12\\ 4y=-3x+12\\ y=-\frac{3}{4}x+3\end{array}$

If the line is parallel to the line $y=-\frac{3}{4}x+3$ , its slope must be $-\frac{3}{4}$ .

The required line has slop

  $\begin{array}{l}{m}_2=-\frac{1}{m_1}=-\left(-\frac{4}{3}\right)\\ m_2=-\frac{1}{m_1}=\frac{4}{3}\end{array}$

Line passes through $\left(8,0\right)$

  $\begin{array}{l}y-0=\frac{4}{3}\left(x-8\right)\\ y=\frac{4}{3}x-\frac{32}{3}\\ 3y=4x-32\\ -4x+3y+32=0\\ -\left(4x-3y-32\right)=0\\ 4x-3y-32=0\end{array}$

Hence, the equation is $4x-3y-32=0$ .