One possible condition for the three points (a,b), (b,a) and `(a^(2),-b^(2))` to be collinear is

Options

A. `a-b=2`
B. `a+b=2`
C. `a=1+b`
D. `a=1-b`

Explanation

To determine one possible condition for the three points \((a,b)\), \((b,a)\), and \((a^2,-b^2)\) to be collinear, we can use the concept of the determinant of a matrix formed by these points. The points are collinear if the determinant of the following matrix is equal to zero:

\[ \begin{vmatrix} a & b & 1 \\ b & a & 1 \\ a^2 & -b^2 & 1 \end{vmatrix} = 0 \]

### Step 1: Set up the determinant
We start with the determinant:

\[ D = \begin{vmatrix} a & b & 1 \\ b & a & 1 \\ a^2 & -b^2 & 1 \end{vmatrix} \]

### Step 2: Expand the determinant
Using the formula for the determinant of a 3x3 matrix, we expand it:

\[ D = a \begin{vmatrix} a & 1 \\ -b^2 & 1 \end{vmatrix} - b \begin{vmatrix} b & 1 \\ a^2 & 1 \end{vmatrix} + 1 \begin{vmatrix} b & a \\ a^2 & -b^2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} a & 1 \\ -b^2 & 1 \end{vmatrix} = a \cdot 1 - (-b^2) \cdot 1 = a + b^2\)

2. \(\begin{vmatrix} b & 1 \\ a^2 & 1 \end{vmatrix} = b \cdot 1 - a^2 \cdot 1 = b - a^2\)

3. \(\begin{vmatrix} b & a \\ a^2 & -b^2 \end{vmatrix} = b \cdot (-b^2) - a \cdot a^2 = -b^3 - a^3\)

### Step 4: Substitute back into the determinant
Now substituting these back into the expression for \(D\):

\[ D = a(a + b^2) - b(b - a^2) + (-b^3 - a^3) \]

### Step 5: Simplify the expression
Expanding and simplifying:

\[ D = a^2 + ab^2 - (b^2 - ba^2) - b^3 - a^3 \]
\[ D = a^2 + ab^2 - b^2 + ba^2 - b^3 - a^3 \]
\[ D = (a^2 + ba^2) + (ab^2 - b^2) - b^3 - a^3 \]
\[ D = a^2(1 + b) + b^2(a - 1) - b^3 - a^3 \]

### Step 6: Set the determinant to zero
Setting \(D = 0\):

\[ a^2(1 + b) + b^2(a - 1) - b^3 - a^3 = 0 \]

### Step 7: Factor the expression
This can be factored or rearranged to find conditions. One possible condition derived from this equation is:

\[ a - b = 0 \quad \text{or} \quad a + b = 0 \quad \text{or} \quad a = 1 + b \]

### Conclusion
Thus, one possible condition for the three points to be collinear is:

\[ a = 1 + b \]