Question

find the slope of a line perpendicular to the line whose equation is $2x + 3y = -3$. fully simplify your answer.

Explanation:

Step1: Find slope of given line

Rewrite \(2x + 3y=-3\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
Subtract \(2x\) from both sides: \(3y=-2x - 3\).
Divide by 3: \(y=-\frac{2}{3}x - 1\).
So slope of given line \(m_1 = -\frac{2}{3}\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\), i.e., \(m_1\times m_2=-1\).
Let \(m_2\) be slope of perpendicular line.
\(-\frac{2}{3}\times m_2=-1\).
Solve for \(m_2\): \(m_2=\frac{-1}{-\frac{2}{3}}=\frac{3}{2}\).

Answer:

\(\frac{3}{2}\)