The points `(-a, -b), (0, 0),(a, b)` and `(a^(2),ab)` are

Options

A. collinear
B. vertices of a rectangle
C. vertices of a parallelogram
D. None of the above

Answer

To determine the relationship between the points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, ab)\), we will calculate the distances between these points and check if they are collinear.

### Step-by-Step Solution:

1. **Identify the Points:**
- Let \( A = (-a, -b) \)
- Let \( B = (0, 0) \)
- Let \( C = (a, b) \)
- Let \( D = (a^2, ab) \)

2. **Calculate the Distance \( AB \):**
- Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
\[ AB = \sqrt{(0 - (-a))^2 + (0 - (-b))^2} = \sqrt{(a)^2 + (b)^2} = \sqrt{a^2 + b^2} \]

3. **Calculate the Distance \( BC \):**
\[ BC = \sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{(a)^2 + (b)^2} = \sqrt{a^2 + b^2} \]

4. **Calculate the Distance \( CD \):**
- For points \( C \) and \( D \):
\[ CD = \sqrt{(a^2 - a)^2 + (ab - b)^2} \]
- Simplifying:
\[ CD = \sqrt{(a^2 - a)^2 + (ab - b)^2} = \sqrt{(a(a - 1))^2 + (b(a - 1))^2} = \sqrt{(a - 1)^2(a^2 + b^2)} \]

5. **Calculate the Distance \( AD \):**
- For points \( A \) and \( D \):
\[ AD = \sqrt{(a^2 - (-a))^2 + (ab - (-b))^2} \]
- Simplifying:
\[ AD = \sqrt{(a^2 + a)^2 + (ab + b)^2} = \sqrt{(a + 1)^2(a^2 + b^2)} \]

6. **Check for Collinearity:**
- For the points to be collinear, the sum of the distances \( AB + BC + CD \) should equal \( AD \):
\[ AB + BC + CD = \sqrt{a^2 + b^2} + \sqrt{a^2 + b^2} + \sqrt{(a - 1)^2(a^2 + b^2)} = 2\sqrt{a^2 + b^2} + \sqrt{(a - 1)^2(a^2 + b^2)} \]
- This should equal \( AD = \sqrt{(a + 1)^2(a^2 + b^2)} \).

7. **Conclusion:**
- After simplifying, we find that the distances are equal, confirming that the points are collinear.

### Final Answer:
The points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, ab)\) are **collinear**.