Using determinants, find the equation of line passing through (0, 3) and (1, 1).

Answer

To find the equation of the line passing through the points (0, 3) and (1, 1) using determinants, we can follow these steps:

### Step 1: Set up the determinant
We use the determinant formula for the equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\). The determinant is set up as follows:

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

For our points \((x_1, y_1) = (0, 3)\) and \((x_2, y_2) = (1, 1)\), we substitute these values into the determinant:

\[ \begin{vmatrix} x & y & 1 \\ 0 & 3 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
Now, we calculate the determinant:

\[ = x \begin{vmatrix} 3 & 1 \\ 1 & 1 \end{vmatrix} - y \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

2. \(\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = (0 \cdot 1) - (1 \cdot 1) = 0 - 1 = -1\)

3. \(\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = (0 \cdot 1) - (3 \cdot 1) = 0 - 3 = -3\)

Substituting these values back into the determinant calculation:

\[ = x(2) - y(-1) + 1(-3) = 2x + y - 3 \]

### Step 3: Set the determinant equal to zero
Now we set the determinant equal to zero:

\[ 2x + y - 3 = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us the equation of the line:

\[ 2x + y = 3 \]

This is the required equation of the line passing through the points (0, 3) and (1, 1).