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### Problem Breakdown:

We are given the equation of a straight line:

\[
x - 2y + 4 = 0
\]

The line cuts the x-axis at point \( A \) and the y-axis at point \( B \). We are tasked with:

1. Finding the coordinates of points \( A \) and \( B \).
2. Using points \( A \), \( B \), and the origin \( O \) to form \( \triangle OAB \), and calculating:
   - (i) The orthocenter of \( \triangle OAB \)
   - (ii) The circumcenter of \( \triangle OAB \)
   - (iii) The centroid of \( \triangle OAB \)

---

### Part (a): Finding the Coordinates of \( A \) and \( B \)

#### 1. Point \( A \) (x-axis intercept):

To find the x-axis intercept, set \( y = 0 \) in the equation of the line:

\[
x - 2(0) + 4 = 0 \implies x + 4 = 0 \implies x = -4
\]

Thus, the coordinates of point \( A \) are \( (-4, 0) \).

#### 2. Point \( B \) (y-axis intercept):

To find the y-axis intercept, set \( x = 0 \) in the equation of the line:

\[
0 - 2y + 4 = 0 \implies -2y + 4 = 0 \implies -2y = -4 \implies y = 2
\]

Thus, the coordinates of point \( B \) are \( (0, 2) \).

### Part (b): Properties of \( \triangle OAB \)

The points we have are:
- \( O(0, 0) \) (the origin),
- \( A(-4, 0) \) (x-axis intercept),
- \( B(0, 2) \) (y-axis intercept).

#### (i) Orthocenter of \( \triangle OAB \)

In a right-angled triangle, the orthocenter is the vertex where the right angle occurs. Since \( \triangle OAB \) is right-angled at \( O \), the orthocenter is the origin itself:
\[
\text{Orthocenter} = O(0, 0)
\]

#### (ii) Circumcenter of \( \triangle OAB \)

The circumcenter of a right-angled triangle is the midpoint of the hypotenuse. The hypotenuse here is \( AB \).

The midpoint of \( A(-4, 0) \) and \( B(0, 2) \) is:

\[
\left( \frac{-4 + 0}{2}, \frac{0 + 2}{2} \right) = (-2, 1)
\]

Thus, the circumcenter is \( (-2, 1) \).

#### (iii) Centroid of \( \triangle OAB \)

The centroid is the average of the three vertices' coordinates. The vertices are \( O(0, 0) \), \( A(-4, 0) \), and \( B(0, 2) \).

The centroid is:

\[
\left( \frac{0 + (-4) + 0}{3}, \frac{0 + 0 + 2}{3} \right) = \left( \frac{-4}{3}, \frac{2}{3} \right)
\]

Thus, the centroid is \( \left( -\frac{4}{3}, \frac{2}{3} \right) \).

---

### Summary of Solutions:

1. Coordinates of \( A \) and \( B \):
   - \( A(-4, 0) \)
   - \( B(0, 2) \)
2. For \( \triangle OAB \):
   - (i) Orthocenter: \( (0, 0) \)
   - (ii) Circumcenter: \( (-2, 1) \)
   - (iii) Centroid: \( \left( -\frac{4}{3}, \frac{2}{3} \right) \)

Would you like more details on any part? 

### Related Questions:
1. How do you find the incenter of a triangle?
2. How do you calculate the area of \( \triangle OAB \) using coordinates?
3. How would you derive the equation of the line perpendicular to \( AB \)?
4. What is the length of the hypotenuse \( AB \) in this triangle?
5. How do you find the equation of the circumcircle of \( \triangle OAB \)?

**Tip**: The orthocenter of a right-angled triangle always lies at the vertex of the right angle!

As shown in the figure, the equation of a straight line is x - 2y + 4 = 0. The line cuts the x-axis at A and the y-axis at B respectively. (a) Find the coordinates of A and B. (b) A, B, and the origin O form ΔOAB. (i) Write down the orthocenter of ΔOAB. (ii) Write down the circumcenter of ΔOAB. (iii) Write down the centroid of ΔOAB.

Learn how to calculate the orthocenter, circumcenter, and centroid of triangle ΔOAB formed by the line x - 2y + 4 = 0. This problem involves analytical geometry concepts such as finding intercepts, centroids, and key points in right-angled triangles. Suitable for Grades 10-12.

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