The ACT Test Math Practice Test Questions | ACT - Free Printable

Problem Analysis:

The image shows a coordinate plane with three points marked:
1. $(-1, -1)$
2. $(2, 1)$
3. $(6, -5)$

The task is to solve a problem involving these points. Since the specific problem is not explicitly stated in the question, I will assume a common type of problem: finding the equation of the line passing through two of the points. If the problem is different, please clarify, and I can adjust the solution accordingly.

Solution:

#### Step 1: Identify the Points
We are given three points:
- $ A = (-1, -1) $
- $ B = (2, 1) $
- $ C = (6, -5) $

To find the equation of a line, we need two points. Let's use points $ A $ and $ B $ to find the equation of the line passing through them.

#### Step 2: Calculate the Slope
The slope $ m $ of a line passing through two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Using points $ A = (-1, -1) $ and $ B = (2, 1) $:
\[
m = \frac{1 - (-1)}{2 - (-1)} = \frac{1 + 1}{2 + 1} = \frac{2}{3}
\]

#### Step 3: Use the Point-Slope Form
The point-slope form of the equation of a line is:
\[
y - y_1 = m(x - x_1)
\]

Using point $ A = (-1, -1) $ and the slope $ m = \frac{2}{3} $:
\[
y - (-1) = \frac{2}{3}(x - (-1))
\]
\[
y + 1 = \frac{2}{3}(x + 1)
\]

#### Step 4: Simplify the Equation
Distribute the slope on the right-hand side:
\[
y + 1 = \frac{2}{3}x + \frac{2}{3}
\]

Subtract 1 from both sides to isolate $ y $:
\[
y = \frac{2}{3}x + \frac{2}{3} - 1
\]

Convert 1 to a fraction with a denominator of 3:
\[
y = \frac{2}{3}x + \frac{2}{3} - \frac{3}{3}
\]
\[
y = \frac{2}{3}x - \frac{1}{3}
\]

#### Step 5: Write the Final Equation
The equation of the line passing through points $ A $ and $ B $ is:
\[
y = \frac{2}{3}x - \frac{1}{3}
\]

Verification:

To verify, we can check if the third point $ C = (6, -5) $ lies on this line. Substitute $ x = 6 $ into the equation:
\[
y = \frac{2}{3}(6) - \frac{1}{3}
\]
\[
y = \frac{12}{3} - \frac{1}{3}
\]
\[
y = \frac{11}{3}
\]

Since $ y = -5 $ does not equal $ \frac{11}{3} $, point $ C $ does not lie on the line passing through $ A $ and $ B $. This confirms that the line we found is specific to points $ A $ and $ B $.

Final Answer:

The equation of the line passing through points $ A = (-1, -1) $ and $ B = (2, 1) $ is:
\[
\boxed{y = \frac{2}{3}x - \frac{1}{3}}
\]