Let's go through the solution step by step.

Part 1: Equation of the Perpendicular Line $l_2$

The equation of the line $l_1$ is given by:

$3x - 2y = 1$

To find the slope of $l_1$, rewrite the equation in the slope-intercept form $y = mx + c$:

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

So, the slope $m_1$ of $l_1$ is $\frac{3}{2}$.

Since $l_2$ is perpendicular to $l_1$, its slope $m_2$ will be the negative reciprocal of $m_1$:

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

We know that line $l_2$ passes through the point $A(4, -1)$. Using the point-slope form of the equation of a line:

$y - y_1 = m_2(x - x_1)$

Substituting $(x_1, y_1) = (4, -1)$ and $m_2 = -\frac{2}{3}$:

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

Simplify this equation:

$y + 1 = -\frac{2}{3}x + \frac{8}{3}$

To eliminate the fractions, multiply the entire equation by 3:

$3y + 3 = -2x + 8$

Rearranging terms to get the equation in the form $ax + by + c = 0$:

$2x + 3y - 5 = 0$

So, the equation of the line $l_2$ is:

$2x + 3y - 5 = 0$

Part 2: Area of Triangle $OPQ$

To find the area of triangle $OPQ$, we need to determine the coordinates of points $P$ and $Q$, where $P$ is the x-intercept and $Q$ is the y-intercept of line $l_1$.

Finding Point $P$ (x-intercept):

At the x-intercept, $y = 0$:

$3x - 2(0) = 1 \implies 3x = 1 \implies x = \frac{1}{3}$

So, the coordinates of $P$ are $\left(\frac{1}{3}, 0\right)$.

Finding Point $Q$ (y-intercept):

At the y-intercept, $x = 0$:

$3(0) - 2y = 1 \implies -2y = 1 \implies y = -\frac{1}{2}$

So, the coordinates of $Q$ are $\left(0, -\frac{1}{2}\right)$.

Area of Triangle $OPQ$:

The area $A$ of triangle $OPQ$ formed by the points $O(0, 0)$, $P\left(\frac{1}{3}, 0\right)$, and $Q\left(0, -\frac{1}{2}\right)$ can be calculated using the formula:

$A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substituting the coordinates $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = \left(\frac{1}{3}, 0\right)$, and $(x_3, y_3) = \left(0, -\frac{1}{2}\right)$:

$A = \frac{1}{2} \left| 0\left(0 - \left(-\frac{1}{2}\right)\right) + \frac{1}{3}\left(-\frac{1}{2} - 0\right) + 0\left(0 - 0\right) \right|$

$A = \frac{1}{2} \left| \frac{1}{3} \times -\frac{1}{2} \right|$

$A = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Thus, the area of triangle $OPQ$ is $\frac{1}{12}$ square units.

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Would you like to explore any specific part of the solution further, or do you have any questions? Here are some related questions to consider:

1. What is the equation of a line parallel to $l_1$ that passes through a different point?
2. Calculate the distance between point $A(4, -1)$ and the line $l_1$.
3. Determine the coordinates of the point where lines $l_1$ and $l_2$ intersect.
4. Find the midpoint of the line segment $PQ$.
5. What is the slope of a line that bisects the angle between $l_1$ and $l_2$?

Tip: When calculating the area of a triangle using its vertices, always double-check the coordinates to avoid mistakes!