We are asked to find the equation of a line that is perpendicular to the line given by the equation:

$3x + 2y = 6$

Step 1: Find the slope of the given line

First, rewrite the given equation $3x + 2y = 6$ in slope-intercept form, $y = mx + b$, where $m$ is the slope.

$2y = -3x + 6$

$y = -\frac{3}{2}x + 3$

The slope of this line is $m_1 = -\frac{3}{2}$.

Step 2: Find the slope of the perpendicular line

The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope. So, the slope of the perpendicular line, $m_2$, is:

$m_2 = \frac{2}{3}$

Step 3: Use the point-slope form to find the equation

The point-slope form of the equation of a line is:

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

Here, $m = \frac{2}{3}$ and the given point is $(6, -3)$, so $x_1 = 6$ and $y_1 = -3$.

Substitute these values into the point-slope form:

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

$y + 3 = \frac{2}{3}(x - 6)$

Step 4: Simplify the equation

Distribute the slope $\frac{2}{3}$:

$y + 3 = \frac{2}{3}x - 4$

Now, subtract 3 from both sides:

$y = \frac{2}{3}x - 7$

Final Answer:

The equation of the line that is perpendicular to $3x + 2y = 6$ and passes through the point $(6, -3)$ is:

$y = \frac{2}{3}x - 7$

Do you want more details on any step or have any questions?

Here are some related questions you might want to explore:

1. What is the equation of a line parallel to $3x + 2y = 6$ passing through $(6, -3)$?
2. How do you derive the point-slope form of a line equation?
3. Can two lines with negative reciprocal slopes ever be parallel?
4. What is the general method to find the slope of a line from an equation in standard form?
5. What are the conditions for two lines to be perpendicular?

Tip: Always remember, the slopes of perpendicular lines are negative reciprocals of each other!