The points `(-a,-b)`, `(0,0)`. `(a,b)` and `(a^(2),a^(3))` are

Options

A. Collinear
B. Vertices of rectangle
C. Vertices of parallelogram
D. Vertices of square

Answer

To determine the relationship among the points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, a^3)\), we will check if they are collinear.

### Step 1: Identify the points
Let:
- Point A = \((-a, -b)\)
- Point B = \((0, 0)\)
- Point C = \((a, b)\)
- Point D = \((a^2, a^3)\)

### Step 2: Find the midpoint of line segment AC
The midpoint \(M\) of line segment \(AC\) can be calculated using the midpoint formula:
\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the coordinates of points A and C:
\[ M = \left(\frac{-a + a}{2}, \frac{-b + b}{2}\right) = \left(0, 0\right) \]
This midpoint is the same as the coordinates of point B.

### Step 3: Check collinearity of points A, B, and C
Since the midpoint of \(AC\) is point \(B\), points \(A\), \(B\), and \(C\) are collinear.

### Step 4: Write the equation of line AC
Using the two-point form of the equation of a line, we can find the equation of line \(AC\):
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b - (-b)}{a - (-a)} = \frac{b + b}{a + a} = \frac{2b}{2a} = \frac{b}{a} \]
Thus, the equation becomes:
\[ y - 0 = \frac{b}{a}(x - 0) \]
or
\[ y = \frac{b}{a}x \]

### Step 5: Substitute point D into the equation
Now we will check if point \(D\) lies on the line \(y = \frac{b}{a}x\). Substitute \(x = a^2\):
\[ y = \frac{b}{a}(a^2) = ab \]
Since the coordinates of point \(D\) are \((a^2, a^3)\), we can see that:
\[ y = ab \quad \text{(from the line equation)} \]
However, point \(D\) has coordinates \((a^2, a^3)\), which means \(y\) should equal \(a^3\).

### Step 6: Conclusion
Since \(ab \neq a^3\) unless \(b = a^2\), point \(D\) does not lie on the line formed by points \(A\), \(B\), and \(C\). Therefore, the four points are not collinear.

### Final Answer
The points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, a^3)\) are **not collinear**.