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

Answer

To determine whether the points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, ab)\) are collinear, we can follow these steps:

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

### Step 2: Calculate the Slope of Line AB
The slope \( m \) between points \( A \) and \( B \) is given by the formula:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-b)}{0 - (-a)} = \frac{b}{a} \]

### Step 3: Write the Equation of Line AB
Using point-slope form, the equation of the line passing through points \( A \) and \( B \) can be expressed as:
\[ y - y_1 = m(x - x_1) \]
Substituting \( B(0, 0) \) and the slope \( m_{AB} \):
\[ y - 0 = \frac{b}{a}(x - 0) \implies y = \frac{b}{a}x \]

### Step 4: Check if Point C Lies on Line AB
Substituting point \( C(a, b) \) into the equation \( y = \frac{b}{a}x \):
\[ b = \frac{b}{a}(a) \]
This simplifies to:
\[ b = b \]
Thus, point \( C \) lies on line \( AB \).

### Step 5: Check if Point D Lies on Line AB
Now, substitute point \( D(a^2, ab) \) into the same line equation:
\[ ab = \frac{b}{a}(a^2) \]
This simplifies to:
\[ ab = b \cdot a \]
Assuming \( b \neq 0 \), we can cancel \( b \) from both sides:
\[ a = a \]
Thus, point \( D \) also lies on line \( AB \).

### Conclusion
Since both points \( C \) and \( D \) lie on the line formed by points \( A \) and \( B \), we conclude that the points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, ab)\) are collinear.

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