The points `(x, y ), (x_1, y_1) and (x - x_1, y - y_1)` are collinear, if

Options

A. `xy_1 = x_1 y`
B. `xy =x_1 y_1`
C. `x x_1 = yy_1`
D. `x + x_1 = y + y_1`

Answer

To determine the condition under which the points \((x, y)\), \((x_1, y_1)\), and \((x - x_1, y - y_1)\) are collinear, we can use the concept of the determinant of a matrix formed by these points. If the determinant is zero, the points are collinear.

### Step-by-Step Solution:

1. **Set up the determinant**:
We can represent the points as follows:
\[ \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x - x_1 & y - y_1 & 1 \end{vmatrix} \]
The determinant of this matrix must equal zero for the points to be collinear.

2. **Calculate the determinant**:
The determinant can be calculated using the formula for a 3x3 matrix:
\[ D = x \begin{vmatrix} y_1 & 1 \\ y - y_1 & 1 \end{vmatrix} - y \begin{vmatrix} x_1 & 1 \\ x - x_1 & 1 \end{vmatrix} + 1 \begin{vmatrix} x_1 & y_1 \\ x - x_1 & y - y_1 \end{vmatrix} \]

3. **Expand the determinants**:
- The first determinant:
\[ \begin{vmatrix} y_1 & 1 \\ y - y_1 & 1 \end{vmatrix} = y_1 \cdot 1 - (y - y_1) \cdot 1 = y_1 - (y - y_1) = 2y_1 - y \]
- The second determinant:
\[ \begin{vmatrix} x_1 & 1 \\ x - x_1 & 1 \end{vmatrix} = x_1 \cdot 1 - (x - x_1) \cdot 1 = x_1 - (x - x_1) = 2x_1 - x \]
- The third determinant:
\[ \begin{vmatrix} x_1 & y_1 \\ x - x_1 & y - y_1 \end{vmatrix} = x_1(y - y_1) - y_1(x - x_1) = x_1y - x_1y_1 - y_1x + y_1x_1 = x_1y - y_1x \]

4. **Combine the results**:
Putting it all together, we have:
\[ D = x(2y_1 - y) - y(2x_1 - x) + (x_1y - y_1x) \]
Simplifying this gives:
\[ D = 2xy_1 - xy - 2yx_1 + yx + x_1y - y_1x \]
Notice that \( -xy + xy \) cancels out, leading to:
\[ D = 2xy_1 - 2yx_1 + x_1y - y_1x \]

5. **Set the determinant to zero**:
For the points to be collinear, we set \( D = 0 \):
\[ 2xy_1 - 2yx_1 + x_1y - y_1x = 0 \]
Rearranging gives:
\[ x_1y = xy_1 \]

### Conclusion:
Thus, the points \((x, y)\), \((x_1, y_1)\), and \((x - x_1, y - y_1)\) are collinear if:
\[ xy_1 = x_1y \]