Question

what is the slope of a line that is perpendicular to the line $y = 8x + 5$? $-8$ $8$ $-\frac{1}{8}$ $\frac{1}{8}$

Explanation:

Step1: Identify the slope of the given line

The equation of the given line is in the slope - intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. For the line \(y = 8x+5\), by comparing with \(y=mx + b\), we can see that the slope \(m_1\) of the given line is \(8\).

Step2: Recall the formula for the slope of a perpendicular line

If two lines with slopes \(m_1\) and \(m_2\) are perpendicular to each other, then the product of their slopes is \(- 1\), i.e., \(m_1\times m_2=-1\). We know \(m_1 = 8\), and we want to find \(m_2\).
From \(m_1\times m_2=-1\), we can solve for \(m_2\) by the formula \(m_2=-\frac{1}{m_1}\).

Step3: Calculate the slope of the perpendicular line

Substitute \(m_1 = 8\) into the formula \(m_2=-\frac{1}{m_1}\). So \(m_2=-\frac{1}{8}\).

Answer:

\(-\frac{1}{8}\) (the option corresponding to \(-\frac{1}{8}\))