What is the length of the line segment AB, given point A is ` (a cos theta, a sin theta )` and point B is ` (a sin theta , - a costheta ) ` ?

Options

A. ` (a)/(sqrt(2))`
B. ` a sqrt""2 `
C. `2 a `
D. None of these

Answer

To find the length of the line segment AB given the points A and B, we will use the distance formula. The coordinates of point A are \( A(a \cos \theta, a \sin \theta) \) and the coordinates of point B are \( B(a \sin \theta, -a \cos \theta) \).

### Step-by-step Solution:

1. **Identify the coordinates of points A and B:**
- Point A: \( (x_1, y_1) = (a \cos \theta, a \sin \theta) \)
- Point B: \( (x_2, y_2) = (a \sin \theta, -a \cos \theta) \)

2. **Apply the distance formula:**
The distance \( AB \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

3. **Substitute the coordinates into the formula:**
\[ AB = \sqrt{(a \sin \theta - a \cos \theta)^2 + (-a \cos \theta - a \sin \theta)^2} \]

4. **Simplify the expressions:**
- For the first term:
\[ (a \sin \theta - a \cos \theta)^2 = a^2 (\sin \theta - \cos \theta)^2 \]
- For the second term:
\[ (-a \cos \theta - a \sin \theta)^2 = a^2 (\cos \theta + \sin \theta)^2 \]

5. **Combine the terms:**
\[ AB = \sqrt{a^2 (\sin \theta - \cos \theta)^2 + a^2 (\sin \theta + \cos \theta)^2} \]
\[ = a \sqrt{(\sin \theta - \cos \theta)^2 + (\sin \theta + \cos \theta)^2} \]

6. **Expand the squares:**
- Expanding \( (\sin \theta - \cos \theta)^2 \):
\[ = \sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta \]
- Expanding \( (\sin \theta + \cos \theta)^2 \):
\[ = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta \]

7. **Combine the expanded terms:**
\[ = (\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta) + (\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta) \]
\[ = 2(\sin^2 \theta + \cos^2 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = 2 \cdot 1 = 2 \]

8. **Final calculation:**
\[ AB = a \sqrt{2} \]

### Conclusion:
The length of the line segment AB is \( a \sqrt{2} \).