The problem asks to find the equation of a line that is perpendicular to the line $2x + 3y = 5$ and passes through the point $(-2, -3)$.

1. Problem Description

The problem asks to find the equation of a line that is perpendicular to the line $2x + 3y = 5$ and passes through the point $(-2, -3)$.

2. Solution Steps

First, we need to find the slope of the given line $2x + 3y = 5$. We can rewrite this equation in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

$3y = -2x + 5$

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

So, the slope of the given line is $m_1 = -\frac{2}{3}$.

Next, we need to find the slope of the line perpendicular to the given line. The slopes of perpendicular lines are negative reciprocals of each other. If $m_1$ is the slope of the first line and $m_2$ is the slope of the perpendicular line, then $m_1 \cdot m_2 = -1$.

$m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}$

So, the slope of the perpendicular line is $m_2 = \frac{3}{2}$.

Now we have the slope of the perpendicular line and a point $(-2, -3)$ that it passes through. We can use the point-slope form of a line equation, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point and $m$ is the slope.

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

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

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

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

Finally, we can rewrite the equation in the standard form $Ax + By = C$.

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

$2y = 3x$

$-3x + 2y = 0$

$3x - 2y = 0$

3. Final Answer

The equation of the line perpendicular to $2x + 3y = 5$ and passing through $(-2, -3)$ is $y = \frac{3}{2}x$ or $3x-2y=0$.