The problem asks us to find the equation of a line that is perpendicular to the line $3x + 2y = 1$ and passes through the point $(-2, 1)$. We need to express the answer in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

1. Problem Description

The problem asks us to find the equation of a line that is perpendicular to the line $3x + 2y = 1$ and passes through the point $(-2, 1)$. We need to express the answer in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

2. Solution Steps

First, we need to find the slope of the given line $3x + 2y = 1$. To do this, we rewrite the equation in slope-intercept form:

$2y = -3x + 1$

$y = -\frac{3}{2}x + \frac{1}{2}$

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

The slope of a line perpendicular to the given line is the negative reciprocal of the given line's slope. Therefore, the slope of the perpendicular line is:

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

Now we know the slope of the perpendicular line is $\frac{2}{3}$, and it passes through the point $(-2, 1)$. We can use the point-slope form of a line to find the equation:

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

$y - 1 = \frac{2}{3}(x - (-2))$

$y - 1 = \frac{2}{3}(x + 2)$

Now we convert this to slope-intercept form:

$y - 1 = \frac{2}{3}x + \frac{4}{3}$

$y = \frac{2}{3}x + \frac{4}{3} + 1$

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

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

3. Final Answer

The equation of the line is $y = \frac{2}{3}x + \frac{7}{3}$.