To find the equation of line \( \ell \), which is parallel to line \( k \) and passes through the point \( (-3, -2) \), we first need to identify the slope of line \( k \).
The equation of line \( k \) is given as:
\[ y = \frac{10}{3} x + 9 \]
From this equation, we can see that the slope \( m \) of line \( k \) is \( \frac{10}{3} \).
Since line \( \ell \) is parallel to line \( k \), it will have the same slope. Therefore, the slope of line \( \ell \) is also \( \frac{10}{3} \).
Now, we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) \) is the point on the line. In this case, the point is \( (-3, -2) \) and the slope \( m \) is \( \frac{10}{3} \).
Substituting in the values, we get:
\[ y - (-2) = \frac{10}{3}(x - (-3)) \]
This simplifies to:
\[ y + 2 = \frac{10}{3}(x + 3) \]
Next, we distribute \( \frac{10}{3} \):
\[ y + 2 = \frac{10}{3}x + \frac{10}{3} \cdot 3 \]
Since \( \frac{10}{3} \cdot 3 = 10 \), we have:
\[ y + 2 = \frac{10}{3}x + 10 \]
Now, we can isolate \( y \) by subtracting 2 from both sides:
\[ y = \frac{10}{3}x + 10 - 2 \]
This simplifies to:
\[ y = \frac{10}{3}x + 8 \]
Thus, the equation of line \( \ell \) in slope-intercept form is:
\[ \boxed{y = \frac{10}{3}x + 8} \]