https://file.aiquickdraw.com/mathbot/file/1725385815448ry331p2y.jpg

The image contains a mathematical problem based on the graphs of two linear equations. The specific tasks are:

1. **Equations provided:**
   \[
   G_1: y = -3x - 3 \quad \text{and} \quad G_2: y = \frac{1}{3}x - 2
   \]

2. **Questions:**
   - **(a)** State the coordinates of the points where each of the graphs intersects:
     - **i.** the x-axis, indicated by \( A \) and \( B \),
     - **ii.** the y-axis, indicated by \( C \) and \( D \).
     - The two graphs intersect at the point \( P \).
   - **(b)** Find the exact area of the triangle \( ABD \).

### Solution Steps:

#### (a) Finding the intersection points:

**i. Intersection with the x-axis:**

For intersection with the x-axis, \( y = 0 \).

- **For \( G_1: y = -3x - 3 \):**
  \[
  0 = -3x - 3
  \]
  \[
  3x = -3 \implies x = -1
  \]
  So, point \( A \) is \( (-1, 0) \).

- **For \( G_2: y = \frac{1}{3}x - 2 \):**
  \[
  0 = \frac{1}{3}x - 2
  \]
  \[
  \frac{1}{3}x = 2 \implies x = 6
  \]
  So, point \( B \) is \( (6, 0) \).

**ii. Intersection with the y-axis:**

For intersection with the y-axis, \( x = 0 \).

- **For \( G_1: y = -3x - 3 \):**
  \[
  y = -3(0) - 3 = -3
  \]
  So, point \( C \) is \( (0, -3) \).

- **For \( G_2: y = \frac{1}{3}x - 2 \):**
  \[
  y = \frac{1}{3}(0) - 2 = -2
  \]
  So, point \( D \) is \( (0, -2) \).

#### (b) Finding the exact area of the triangle \( ABD \):

The vertices of the triangle \( ABD \) are \( A(-1, 0) \), \( B(6, 0) \), and \( D(0, -2) \).

Using the formula for the area of a triangle given three vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
\]
Substituting the points \( A(-1, 0) \), \( B(6, 0) \), and \( D(0, -2) \):
\[
\text{Area} = \frac{1}{2} \left| -1(0 - (-2)) + 6(-2 - 0) + 0(0 - 0) \right|
\]
\[
= \frac{1}{2} \left| -1(2) + 6(-2) + 0 \right|
\]
\[
= \frac{1}{2} \left| -2 - 12 \right|
\]
\[
= \frac{1}{2} \times 14 = 7 \, \text{square units}
\]

### Final Answers:
1. **Points of Intersection:**
   - \( A(-1, 0) \)
   - \( B(6, 0) \)
   - \( C(0, -3) \)
   - \( D(0, -2) \)

2. **Area of Triangle \( ABD \):** \( 7 \) square units.

Would you like more details on any of these calculations or explanations?  

Here are some related questions to consider:
1. What is the general process for finding the intersection of two lines?
2. How does changing the slope of a line affect its intersection points with the axes?
3. Can you determine the area of a triangle formed by any three points on a graph?
4. How would the area calculation change if the coordinates of point \( D \) were different?
5. What are other methods to find the area of a triangle if you know the vertices?

**Tip:** When finding where a graph intersects the x-axis or y-axis, remember that the value of the other coordinate (y or x, respectively) is zero.

The figure above shows the graphs of G1: y = -3x - 3 and G2: y = 1/3x - 2. 
a) State the coordinates of the points where each of i) the x-axis, indicated by A and B, ii) the y-axis, indicated by C and D. The two graphs intersect at the point P. 
b) Find the exact area of the triangle ABD.

Learn how to find the intersection points of two linear equations and calculate the area of the triangle formed by these points. This problem covers key algebraic and geometric concepts suitable for students in Grades 9-11.

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