Chapter 5.6, Problem 20P

Write an equation of the line that passes through the given point and is perpendicular to the graph of the given equation.

$(-8,3);y+4=-\frac{2}{3}(x-2)$

To determine

To find: The equation of the line that passes through $(-8,3)$ and is perpendicular to $y+4=-\frac{2}{3}(x-2)$ .

Answer to Problem 20P

The equation of the required line is $y=\frac{3}{2}x+15$ .

Explanation of Solution

Given information:

  $(-8,3);y+4=-\frac{2}{3}(x-2)$

The given equation $y+4=-\frac{2}{3}(x-2)$ is in the point-slope form. Notice that, the slope of the line $y+4=-\frac{2}{3}(x-2)$ is $-\frac{2}{3}$ .

Since the required line is perpendicular to the line $y+4=-\frac{2}{3}(x-2)$ , the slope of the line would be the negative reciprocal of $-\frac{2}{3}$ . That is $-\frac{1}{\left(-\frac{2}{3}\right)}=\frac{3}{2}$ .

So using the point-slope form of a line, the equation of the line can be determined as follows:

  $\begin{array}{l}y-y_1=m(x-x_1)\\ \Rightarrow y-3=\frac{3}{2}(x-(-8))=\frac{3}{2}(x+8)\\ \Rightarrow y-3=\frac{3}{2}x+12\\ \Rightarrow y=\frac{3}{2}x+15\end{array}$

Conclusion:

Therefore, the equation of the required line is $y=\frac{3}{2}x+15$ .