Question

opes of parallel/perpendicular lines
score: 0/2 penalty: 1 off
question
find the slope of a line perpendicular to the line whose equation is 6x + 8y = -128.
fully simplify your answer.
answer attempt 1 out of 2

Explanation:

Step1: Convert the given equation to slope - intercept form ($y = mx + b$)

Starting with $6x + 8y=-128$.
Subtract $6x$ from both sides: $8y=-6x - 128$.
Divide each term by $8$: $y=-\frac{6}{8}x-\frac{128}{8}$, which simplifies to $y =-\frac{3}{4}x - 16$.
The slope of the given line ($m_1$) is $-\frac{3}{4}$.

Step2: Use the perpendicular slope formula ($m_2=-\frac{1}{m_1}$)

If $m_1 =-\frac{3}{4}$, then $m_2=-\frac{1}{-\frac{3}{4}}$.
When we divide by a fraction, we multiply by its reciprocal: $m_2=\frac{4}{3}$.

Answer:

$\frac{4}{3}$