Chapter 5.6, Problem 2STP

For Exercises $1-5,$ choose the correct letter.

Which equation has a graph perpendicular to the graph of $7x=14y-8?$

F. $y=-2x-7$

G. $y=-\frac{1}{2}x+4$

H. $y=\frac{1}{2}x-1$

I. $y=2x+9$

To determine

To find: The equation that has a graph perpendicular to the graph of $7x=14y-8$ from the given options.

Answer to Problem 2STP

The equation that has a graph perpendicular to the graph of $7x=14y-8$ is given in option F.

Explanation of Solution

Given information:

The options given are

  

To determine the equation that has a graph perpendicular to the graph of $7x=14y-8$ , express the equation in slope-intercept form as follows:

  $\begin{array}{l}7x=14y-8\\ \Rightarrow 14y-8=7x\\ \Rightarrow 14y=7x+8\\ \Rightarrow y=\frac{7}{14}x+\frac{8}{14}\\ \Rightarrow y=\frac{1}{2}x+\frac{4}{7}\end{array}$

Here, the slope of the line is $\frac{1}{2}$ . So to find the equation that has a graph perpendicular to the graph of $y=\frac{7}{14}x+\frac{8}{14}$ , we need to look for the equation that has the same slope $-\frac{1}{\left(\frac{1}{2}\right)}=-2$ .

Conclusion:

Among the given equations, the only equation that has the same slope $-2$ is $y=-2x-7$ . Therefore, the equation that has a graph perpendicular to the graph of $7x=14y-8$ is given in option F.