Let's solve question 3 using the given graph for point $C(3, 4)$, where we need to find:

1. The equation of line $\overline{AB}$.
2. The area of $\triangle ABO$.

Step 1: Equation of line $\overline{AB}$

From the graph:

• $A(0, 0)$ is the origin.
• $B(6, 0)$ lies on the x-axis.

The equation of a straight line can be written as: $y = mx + c$ where:

• $m$ is the slope.
• $c$ is the y-intercept.

Slope $m$ of $\overline{AB}$:

The slope between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Substitute: $m = \frac{0 - 0}{6 - 0} = 0$

Since the slope $m = 0$, line $\overline{AB}$ is horizontal. The equation of a horizontal line is simply: $y = c$ Here, $c = 0$, so: $y = 0$

Thus, the equation of $\overline{AB}$ is: $y = 0$

Step 2: Area of $\triangle ABO$

The vertices of $\triangle ABO$ are:

• $A(0, 0)$
• $B(6, 0)$
• $O(3, 4)$

The area of a triangle given its vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substitute the coordinates: $x_1 = 0, \, y_1 = 0, \, x_2 = 6, \, y_2 = 0, \, x_3 = 3, \, y_3 = 4$

$\text{Area} = \frac{1}{2} \left| 0(0 - 4) + 6(4 - 0) + 3(0 - 0) \right|$ $\text{Area} = \frac{1}{2} \left| 0 + 24 + 0 \right|$ $\text{Area} = \frac{1}{2} \times 24 = 12$

Thus, the area of $\triangle ABO$ is: $\boxed{12 \, \text{square units}}$

Final Answers:

1. The equation of $\overline{AB}$ is $\mathbf{y = 0}$.
2. The area of $\triangle ABO$ is $\mathbf{12 \, \text{square units}}$.

Would you like a deeper explanation of any step or further examples?

Follow-Up Questions:

1. What is the general formula for the slope of a line?
2. How do you calculate the area of a triangle using determinants?
3. Can we find the length of side $\overline{AB}$? If yes, how?
4. What happens if point $O$ moves? How does the area of the triangle change?
5. Can you derive the equation of $\overline{AB}$ if the slope is non-zero?

Tip:

Always double-check your slope and intercept calculations, as they directly influence the equation of the line.