Given the points \( P(9, 0) \) and \( Q(0, -12) \), we can calculate several things like the **distance** between the points, the **slope** of the line that passes through them, and the **equation of the line** connecting them.

### 1. Slope of the Line
The formula for the slope \( m \) between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Substituting the given values \( P(9, 0) \) and \( Q(0, -12) \):

\[
m = \frac{-12 - 0}{0 - 9} = \frac{-12}{-9} = \frac{4}{3}
\]

### 2. Equation of the Line
The point-slope form of a line's equation is:

\[
y - y_1 = m(x - x_1)
\]

Using point \( P(9, 0) \) and the slope \( m = \frac{4}{3} \):

\[
y - 0 = \frac{4}{3}(x - 9)
\]

Simplifying:

\[
y = \frac{4}{3}(x - 9)
\]

Expanding:

\[
y = \frac{4}{3}x - 12
\]

Thus, the equation of the line is:

\[
y = \frac{4}{3}x - 12
\]

### 3. Distance Between the Points
The distance \( d \) between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is given by the formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Substituting the values \( P(9, 0) \) and \( Q(0, -12) \):

\[
d = \sqrt{(0 - 9)^2 + (-12 - 0)^2} = \sqrt{(-9)^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15
\]

So, the distance between \( P \) and \( Q \) is 15 units.

Would you like further details or clarification on any of these steps?

Here are 5 related questions:
1. How do you find the midpoint between two points?
2. What is the general form of the equation of a line?
3. How do you determine if two lines are parallel or perpendicular?
4. Can the slope formula be used for vertical lines?
5. How can the distance formula be derived from the Pythagorean theorem?

**Tip:** The slope tells you how steep a line is, and a positive slope means the line rises from left to right.

Explore the step-by-step solution to find the slope, distance, and equation of a line connecting points P(9, 0) and Q(0, -12). Learn key concepts in coordinate geometry, including the slope formula, distance calculation, and point-slope form of a line. Suitable for students in Grades 9-10.

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