If O is the origin and `P(x_(1),y_(1)), Q(x_(2),y_(2))` are two points then `POxxOQ sin angle POQ=`

Options

A. `x_(1)y_(2)+x_(1)y_(2)`
B. `x_(1)y_(2)+x_(2)y_(1)`
C. `|x_(1)y_(2)-x_(2)y_(1)|`
D. none of these

Explanation

To solve the problem, we need to find the expression \( PO \times OQ \times \sin(\angle POQ) \) where \( O \) is the origin, and \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) are two points in the Cartesian coordinate system.

### Step-by-Step Solution:

1. **Identify Points and Distances**:
- The distance \( PO \) from the origin \( O(0, 0) \) to point \( P(x_1, y_1) \) is given by the formula:
\[ PO = \sqrt{x_1^2 + y_1^2} \]
- Similarly, the distance \( OQ \) from the origin \( O(0, 0) \) to point \( Q(x_2, y_2) \) is:
\[ OQ = \sqrt{x_2^2 + y_2^2} \]

2. **Area of Triangle POQ**:
- The area \( A \) of triangle \( POQ \) can be calculated using the determinant formula:
\[ A = \frac{1}{2} \left| x_1y_2 - x_2y_1 \right| \]

3. **Using the Formula for Area**:
- The area of triangle \( POQ \) can also be expressed in terms of the sides and the sine of the angle between them:
\[ A = \frac{1}{2} \cdot PO \cdot OQ \cdot \sin(\angle POQ) \]

4. **Equating the Two Expressions for Area**:
- From the two expressions for the area, we can set them equal to each other:
\[ \frac{1}{2} \left| x_1y_2 - x_2y_1 \right| = \frac{1}{2} \cdot PO \cdot OQ \cdot \sin(\angle POQ) \]

5. **Simplifying the Equation**:
- Cancel the \( \frac{1}{2} \) from both sides:
\[ \left| x_1y_2 - x_2y_1 \right| = PO \cdot OQ \cdot \sin(\angle POQ) \]

6. **Final Result**:
- Therefore, we can conclude that:
\[ PO \cdot OQ \cdot \sin(\angle POQ) = \left| x_1y_2 - x_2y_1 \right| \]