Let's break down and solve the two parts of the problem in the image you uploaded.

(f) Show that the points $X(2, 3)$, $Y(6, -3)$, and $Z(4, 0)$ are collinear.

To prove that these points are collinear, we need to show that the slope between any two pairs of these points is the same.

1. Slope of $XY$: The formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $m = \frac{y_2 - y_1}{x_2 - x_1}$ For points $X(2, 3)$ and $Y(6, -3)$: $m_{XY} = \frac{-3 - 3}{6 - 2} = \frac{-6}{4} = -\frac{3}{2}$

2. Slope of $YZ$: Now, for points $Y(6, -3)$ and $Z(4, 0)$: $m_{YZ} = \frac{0 - (-3)}{4 - 6} = \frac{3}{-2} = -\frac{3}{2}$

3. Slope of $XZ$: Lastly, for points $X(2, 3)$ and $Z(4, 0)$: $m_{XZ} = \frac{0 - 3}{4 - 2} = \frac{-3}{2} = -\frac{3}{2}$

Since the slopes $m_{XY} = m_{YZ} = m_{XZ} = -\frac{3}{2}$, the points are collinear.

(g) Determine the value of $x$ in the circle problem.

In the given diagram, point $O$ is the center of the circle, and we are given:

• $\angle AOC = 120^\circ$
• $\angle ABC = 2x - 4^\circ$

Since $\angle AOC$ is the central angle and subtends arc $AC$, and $\angle ABC$ is an inscribed angle subtended by the same arc, we know that the inscribed angle is half the measure of the central angle.

Thus, the relationship is: $\angle ABC = \frac{1}{2} \times \angle AOC$ Substitute the known values: $2x - 4 = \frac{1}{2} \times 120$ $2x - 4 = 60$ Solve for $x$: $2x = 60 + 4 = 64$ $x = \frac{64}{2} = 32$

Thus, the value of $x$ is $\boxed{32}$.

Would you like more details on any part of the solution?

Here are 5 related questions you might find interesting:

1. How do you find the area of a triangle given three points?
2. How do you determine if three points form a right triangle?
3. What is the difference between a central angle and an inscribed angle in a circle?
4. How can you find the radius of a circle given an angle and a chord?
5. How do you prove that four points form a cyclic quadrilateral?

Tip: For solving geometry problems involving circles, remember that inscribed angles are always half of the central angle that subtends the same arc.