Find the value of `alpha`, if the points (5,1), (-2,-3) and `(8,2 alpha)` are collinear.

Answer

To find the value of `alpha` such that the points (5, 1), (-2, -3), and (8, 2α) are collinear, we can use the determinant method for collinearity. The points are collinear if the area of the triangle formed by them is zero. This can be represented using the following determinant:

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

Where:
- \( (x_1, y_1) = (5, 1) \)
- \( (x_2, y_2) = (-2, -3) \)
- \( (x_3, y_3) = (8, 2α) \)

### Step 1: Set up the determinant

We can set up the determinant as follows:

\[ \begin{vmatrix} 5 & 1 & 1 \\ -2 & -3 & 1 \\ 8 & 2α & 1 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant

Now, we will calculate the determinant:

\[ = 5 \begin{vmatrix} -3 & 1 \\ 2α & 1 \end{vmatrix} - 1 \begin{vmatrix} -2 & 1 \\ 8 & 1 \end{vmatrix} + 1 \begin{vmatrix} -2 & -3 \\ 8 & 2α \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

2. \( \begin{vmatrix} -2 & 1 \\ 8 & 1 \end{vmatrix} = (-2)(1) - (1)(8) = -2 - 8 = -10 \)

3. \( \begin{vmatrix} -2 & -3 \\ 8 & 2α \end{vmatrix} = (-2)(2α) - (-3)(8) = -4α + 24 \)

### Step 3: Substitute back into the determinant equation

Now substituting these back into the determinant equation:

\[ 5(-3 - 2α) - (-10) + (-4α + 24) = 0 \]

This simplifies to:

\[ -15 - 10α + 10 - 4α + 24 = 0 \]

Combining like terms:

\[ -14α + 19 = 0 \]

### Step 4: Solve for α

Now, we can solve for α:

\[ -14α = -19 \]
\[ α = \frac{19}{14} \]

### Final Answer

The value of \( α \) is \( \frac{19}{14} \).

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