The problem asks us to find the equation of a line that passes through the point $(3, 3)$ and is perpendicular to the line $5x - 6y = 4$.

1. Problem Description

The problem asks us to find the equation of a line that passes through the point $(3, 3)$ and is perpendicular to the line $5x - 6y = 4$.

2. Solution Steps

First, we need to find the slope of the given line $5x - 6y = 4$. To do this, we can rewrite the equation in slope-intercept form, $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

$5x - 6y = 4$

$-6y = -5x + 4$

$y = \frac{-5}{-6}x + \frac{4}{-6}$

$y = \frac{5}{6}x - \frac{2}{3}$

So, the slope of the given line is $m_1 = \frac{5}{6}$.

Since we want a line that is perpendicular to the given line, the slope of the new line, $m_2$, must be the negative reciprocal of $m_1$.

$m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{5}{6}} = -\frac{6}{5}$

Now we know the slope of the new line, $m_2 = -\frac{6}{5}$, and a point it passes through, $(3, 3)$. We can use the point-slope form of a linear equation, which is:

$y - y_1 = m(x - x_1)$

where $(x_1, y_1)$ is the given point. Plugging in our values, we have:

$y - 3 = -\frac{6}{5}(x - 3)$

$y - 3 = -\frac{6}{5}x + \frac{18}{5}$

$y = -\frac{6}{5}x + \frac{18}{5} + 3$

$y = -\frac{6}{5}x + \frac{18}{5} + \frac{15}{5}$

$y = -\frac{6}{5}x + \frac{33}{5}$

3. Final Answer

$y = -\frac{6}{5}x + \frac{33}{5}$