Question: The line $l_1$ has the equation $x - 5y - 30 = 0$. The line $l_2$ is parallel to $l_1$ and passes through the point $A (4, 6)$ as shown in the diagram below.
$l_2$ and $l_1$ are parallel diagonal lines with positive slope on a coordinate plane; $l_2$ passes through point $A$ near the upper center-left, and $l_1$ crosses the y-axis below the origin and the x-axis to the right. position\_hint: center
Determine the equation of $l_2$.
Give your answer in the form $ax + by = c$, where $a$, $b$ and $c$ are integers.
1. **State the problem:** We are given the equation of line $l_1$ as $$x - 5y - 30 = 0$$ and a point $A(4,6)$ through which a line $l_2$ parallel to $l_1$ passes. We need to find the equation of $l_2$ in the form $$ax + by = c$$ where $a$, $b$, and $c$ are integers.
2. **Recall the rule for parallel lines:** Parallel lines have the same slope. The slope of $l_1$ can be found by rewriting its equation in slope-intercept form $y = mx + b$.
3. **Find the slope of $l_1$:**
$$x - 5y - 30 = 0 \implies -5y = -x + 30 \implies y = \frac{-x + 30}{-5} = \frac{x}{5} - 6$$
So the slope $m$ of $l_1$ is $$m = \frac{1}{5}$$.
4. **Equation of $l_2$:** Since $l_2$ is parallel to $l_1$, it has the same slope $m = \frac{1}{5}$. Using point-slope form with point $A(4,6)$:
$$y - y_1 = m(x - x_1)$$
$$y - 6 = \frac{1}{5}(x - 4)$$
5. **Simplify the equation:**
$$y - 6 = \frac{1}{5}x - \frac{4}{5}$$
$$y = \frac{1}{5}x - \frac{4}{5} + 6$$
$$y = \frac{1}{5}x + \frac{26}{5}$$
6. **Convert to standard form $ax + by = c$ with integers:** Multiply both sides by 5 to clear denominators:
$$5y = x + 26$$
Rewrite:
$$x - 5y = -26$$
7. **Final answer:** The equation of line $l_2$ is
$$\boxed{x - 5y = -26}$$