The line $l_1$ has equation $3x + 5y - 2 = 0$ - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 2

Since line $l_2$ is perpendicular to line $l_1$, we first need to find the gradient of $l_2$. The gradient of a line perpendicular to another is the negative reciprocal of the original line's gradient.

Therefore, if the gradient of $l_1$ is $-\frac{3}{5}$, the gradient $m$ of $l_2$ is:

$m = -\frac{1}{(-\frac{3}{5})} = \frac{5}{3}$

Next, we use the point $(3, 1)$, which lies on line $l_2$, to find the constant $c$ in the equation of the line:

Using the point-slope form $y - y_1 = m(x - x_1)$, we get:

$1 - y = \frac{5}{3}(3 - x)$

Substituting $(x_1, y_1) = (3, 1)$ gives us:

$y - 1 = \frac{5}{3}(3 - x)$

After rearranging, we get:

$y = \frac{5}{3}(3) - \frac{5}{3}x + 1$

Simplifying further:

$y = 5 - \frac{5}{3}x + 1$

Combining the constants:

$y = - \frac{5}{3}x + 6$

Thus, the equation of line $l_2$ is:

$y = -\frac{5}{3}x + 6$