Question

find the slope of a line perpendicular to the line whose equation is $3x - 3y = 18$. fully simplify your answer.

Explanation:

Step1: Rewrite the given equation in slope - intercept form (\(y = mx + b\), where \(m\) is the slope)

Start with the equation \(3x-3y = 18\).
Subtract \(3x\) from both sides: \(-3y=-3x + 18\).
Divide each term by \(-3\): \(y=\frac{-3x}{-3}+\frac{18}{-3}\).
Simplify to get \(y = x-6\).
So the slope (\(m_1\)) of the given line is \(1\) (since in \(y = mx + b\), \(m = 1\) here).

Step2: Find the slope of the perpendicular line

The slope of a line perpendicular to a line with slope \(m_1\) is the negative reciprocal of \(m_1\). That is, if the slope of one line is \(m_1\), the slope of the line perpendicular to it (\(m_2\)) is \(m_2=-\frac{1}{m_1}\) (when \(m_1
eq0\)).
Since \(m_1 = 1\), the negative reciprocal of \(1\) is \(-\frac{1}{1}=-1\).

Answer:

\(-1\)