The points `(a+1,1), (2a+1,3)` and `(2a+2,2a)` are collinear if

Options

A. `a=-1, 2`
B. `a=(1)/(2), 2`
C. a = 2, 1
D. `a=-(1)/(2), 2`

Answer

To determine if the points \((a+1, 1)\), \((2a+1, 3)\), and \((2a+2, 2a)\) are collinear, we can use the condition that the area formed by these three points must be zero. This can be expressed using the determinant of the following matrix:

\[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]

### Step 1: Identify the points
Let:
- \( (x_1, y_1) = (a+1, 1) \)
- \( (x_2, y_2) = (2a+1, 3) \)
- \( (x_3, y_3) = (2a+2, 2a) \)

### Step 2: Set up the determinant
We set up the determinant as follows:

\[ \begin{vmatrix} a+1 & 1 & 1 \\ 2a+1 & 3 & 1 \\ 2a+2 & 2a & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we expand it using the first row:

\[ = (a+1) \begin{vmatrix} 3 & 1 \\ 2a & 1 \end{vmatrix} - 1 \begin{vmatrix} 2a+1 & 1 \\ 2a+2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2a+1 & 3 \\ 2a+2 & 2a \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 3 & 1 \\ 2a & 1 \end{vmatrix} = 3 \cdot 1 - 1 \cdot 2a = 3 - 2a\)

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

3. \(\begin{vmatrix} 2a+1 & 3 \\ 2a+2 & 2a \end{vmatrix} = (2a+1)(2a) - (3)(2a+2) = 4a^2 + 2a - 6a - 6 = 4a^2 - 4a - 6\)

Putting it all together:

\[ (a+1)(3 - 2a) - (-1) + (4a^2 - 4a - 6) = 0 \]

### Step 4: Simplify the equation
Expanding the equation:

\[ (a+1)(3 - 2a) + 1 + 4a^2 - 4a - 6 = 0 \]

This expands to:

\[ 3a + 3 - 2a^2 - 2a + 1 + 4a^2 - 4a - 6 = 0 \]

Combining like terms:

\[ (4a^2 - 2a^2) + (3a - 2a - 4a) + (3 + 1 - 6) = 0 \]

This simplifies to:

\[ 2a^2 - 3a - 2 = 0 \]

### Step 5: Solve the quadratic equation
Now, we can factor or use the quadratic formula to solve for \(a\):

Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 2\), \(b = -3\), and \(c = -2\):

\[ a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]

Calculating the discriminant:

\[ = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \]

This gives us two solutions:

1. \(a = \frac{8}{4} = 2\)
2. \(a = \frac{-2}{4} = -\frac{1}{2}\)

### Final Result
The points are collinear if \(a = 2\) or \(a = -\frac{1}{2}\).