Question

find the slope of a line perpendicular to the line whose equation is 6x + 8y = -128. fully simplify your answer.

Explanation:

Step1: Rewrite the given line in slope - intercept form

The slope - intercept form is $y = mx + b$, where $m$ is the slope. Starting with $6x + 8y=-128$, we solve for $y$. First, subtract $6x$ from both sides: $8y=-6x - 128$. Then divide each term by 8: $y=-\frac{6}{8}x-\frac{128}{8}=-\frac{3}{4}x - 16$. The slope of this line, $m_1=-\frac{3}{4}$.

Step2: Use the perpendicular - slope relationship

If two lines are perpendicular, the product of their slopes is $- 1$, i.e., $m_1\times m_2=-1$. We know $m_1 = -\frac{3}{4}$, and we want to find $m_2$. Substituting $m_1$ into the formula $-\frac{3}{4}\times m_2=-1$. Solving for $m_2$, we multiply both sides by $-\frac{4}{3}$, so $m_2=\frac{4}{3}$.

Answer:

$\frac{4}{3}$