Chapter 13.7, Problem 27WE

To determine

To find : an equation of line described

Answer to Problem 27WE

The equation is $5y+8y+31=0$

Explanation of Solution

Given information :

Line through $\left(-3,-2\right)$ and perpendicular to the line $8x-5y=0$

Formula used :

  $y-y_1=m\left(x-x_1\right)$

Calculation :

The line

  $\begin{array}{l}8x-5y=0\\ -5y=-8x+0\\ y=\frac{8}{5}x+0\end{array}$

The line is perpendicular to the line $y=\frac{8}{5}x+0$ .

The required line has slope $\frac{-1}{m}=\frac{-1}{\frac{8}{5}}=-\frac{5}{8}$

Line passes through $\left(-3,-2\right)$

  $\begin{array}{l}y-\left(-2\right)=-\frac{5}{8}\left(x-\left(-3\right)\right)\\ y+2=-\frac{5}{8}x-\frac{15}{8}\\ y=-\frac{5}{8}x-\frac{15}{8}-2\\ y=-\frac{5}{8}x-\left(\frac{15+16}{8}\right)\\ y=-\frac{5}{8}x-\frac{31}{8}\\ 8y=-5x-31\\ 5y+8y+31=0\end{array}$

Hence, the equation is $5y+8y+31=0$ .