Using determinants, find the equation of the line joining the points (1,2) and (3,6):

Options

A. y=2x
B. x=3y
C. y=x
D. 4x-y=5

Answer

To find the equation of the line joining the points (1, 2) and (3, 6) using determinants, we can follow these steps:

### Step 1: Set up the determinant
We will use the formula for the area of a triangle formed by three points. The area is given by:

\[ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

For our case, let the points be:
- \( A(1, 2) \)
- \( B(3, 6) \)
- \( P(x, y) \)

Thus, we set up the determinant as follows:

\[ \text{Area} = \frac{1}{2} \begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{vmatrix} \]

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

\[ \begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{vmatrix} = 1 \begin{vmatrix} 6 & 1 \\ y & 1 \end{vmatrix} - 2 \begin{vmatrix} 3 & 1 \\ x & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & 6 \\ x & y \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 6 & 1 \\ y & 1 \end{vmatrix} = 6 \cdot 1 - 1 \cdot y = 6 - y \)

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

3. \( \begin{vmatrix} 3 & 6 \\ x & y \end{vmatrix} = 3y - 6x \)

Putting it all together:

\[ \begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{vmatrix} = 1(6 - y) - 2(3 - x) + 1(3y - 6x) \]

Expanding this gives:

\[ = 6 - y - 6 + 2x + 3y - 6x = -4x + 2y \]

### Step 3: Set the area to zero
Since the points are collinear, the area must be zero:

\[ \frac{1}{2}(-4x + 2y) = 0 \]

This simplifies to:

\[ -4x + 2y = 0 \]

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

\[ 2y = 4x \quad \Rightarrow \quad y = 2x \]

### Final Equation
Thus, the equation of the line joining the points (1, 2) and (3, 6) is:

\[ y = 2x \]