Chapter 13.2, Problem 11PSB

To determine

To show: That $-\frac{a}{b}$ is the slope of the graph of $ax+by+c=0$ .

Answer to Problem 11PSB

It is proved that the slope of the equation $ax+by+c=0$ is $-\frac{a}{b}$ .

Explanation of Solution

Given:

The equations is $ax+by+c=0$ .

Calculation:

The slope intercepts form of equation is $y=mx+b$ , where $m$ is slope and $b$ is y-intercept.

Write the equation $ax+by+c=0$ in $y=mx+b$ form.

  $\begin{array}{c}ax+by+c=0\\ by=-ax-c\\ y=-\frac{a}{b}x-\frac{c}{a}\end{array}$

Compare the equation $y=-\frac{a}{b}x-\frac{c}{a}$ with $y=mx+b$ .

  $m=-\frac{a}{b}$

Therefore, it is proved that the slope of the equation $ax+by+c=0$ is $-\frac{a}{b}$ .