Answer:
To find the equation of a line perpendicular to \(y = \frac{1}{3}x + 27\) and passing through the point (11, -5), we'll use the fact that the slopes of perpendicular lines are negative reciprocals.
The given line has a slope \(m = \frac{1}{3}\). The slope of a line perpendicular to this is \(-3\) (negative reciprocal).
Now, using the point-slope form of a line \((y - y_1) = m(x - x_1)\) and substituting the values:
\[ (y - (-5)) = -3(x - 11) \]
Simplify to get the equation in slope-intercept form:
\[ y + 5 = -3x + 33 \]
Now, solve for \(y\):
\[ y = -3x + 28 \]
So, the equation of the line that passes through (11, -5) and is perpendicular to \(y = \frac{1}{3}x + 27\) is \(y = -3x + 28\).
