To solve for the equation of line $q$, we need to analyze the given information step-by-step.

Step 1: Determine the slope of line $p$

The equation of line $p$ is given in point-slope form: $y - 3 = \frac{1}{4}(x + 1)$ To identify the slope, we recognize that the coefficient of $(x + 1)$ is the slope of line $p$, which is: $m_p = \frac{1}{4}$

Step 2: Determine the slope of line $q$

Line $q$ is perpendicular to line $p$, which means its slope will be the negative reciprocal of the slope of $p$. Thus: $m_q = -\frac{1}{\left(\frac{1}{4}\right)} = -4$

Step 3: Use the point-slope form to find the equation of line $q$

We know that line $q$ passes through the point $(3, -6)$ and has a slope of $-4$. Using the point-slope form $y - y_1 = m(x - x_1)$, we substitute $m = -4$, $x_1 = 3$, and $y_1 = -6$: $y - (-6) = -4(x - 3)$ This simplifies to: $y + 6 = -4(x - 3)$ Expanding and solving for $y$ in slope-intercept form: $y + 6 = -4x + 12$ $y = -4x + 6$

Final Answer

The equation of line $q$ in slope-intercept form is: $y = -4x + 6$

Would you like further clarification on any part of this solution?

Here are some related questions to extend this concept:

1. What would the equation be if line $q$ were parallel to line $p$ instead of perpendicular?
2. How would the equation change if line $q$ passed through a different point, such as $(5, -2)$?
3. What is the interpretation of the y-intercept in the equation of line $q$?
4. How can we convert the slope-intercept form of line $q$ back to point-slope form?
5. What would be the effect on the equation if the slope of line $p$ was negative initially?

Tip: For perpendicular lines, always remember that their slopes multiply to $-1$ because they are negative reciprocals of each other.