Let A be the image of (2, -1) with respect to Y - axis Without transforming the oringin, coordinate axis are turned at an angle `45^(@)` in the clockwise direction. Then, the coordiates of A in the new system are

Options

A. `(-(1)/(sqrt(2)),-(3)/(sqrt(2)))`
B. `(-(3)/(sqrt(2)),-(1)/(sqrt(2)))`
C. `((1)/(sqrt(2)),(3)/(sqrt(2)))`
D. `((3)/(sqrt(2)),(1)/(sqrt(2)))`

Answer

To solve the problem step by step, we will follow the instructions given in the question.

### Step-by-Step Solution:

1. **Identify the Original Coordinates**:
The original coordinates of point A are given as \( (2, -1) \).

2. **Find the Image with Respect to the Y-Axis**:
The image of a point \( (x, y) \) with respect to the Y-axis is given by \( (-x, y) \).
Therefore, the image of point A with respect to the Y-axis is:
\[ A' = (-2, -1) \]

3. **Rotate the Coordinate System**:
We need to rotate the coordinate axes by \( 45^\circ \) in the clockwise direction.
The transformation for rotating a point \( (x, y) \) by \( \theta \) degrees clockwise is given by:
\[ x' = x \cos(\theta) + y \sin(\theta) \]
\[ y' = -x \sin(\theta) + y \cos(\theta) \]
Here, \( \theta = 45^\circ \), so \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \) and \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \).

4. **Substituting the Values**:
Now substituting \( A' = (-2, -1) \) into the transformation equations:
\[ x' = -2 \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{\sqrt{2}} = \frac{-2 - 1}{\sqrt{2}} = \frac{-3}{\sqrt{2}} \]
\[ y' = -(-2) \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{\sqrt{2}} = \frac{2 - 1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

5. **Final Coordinates in the New System**:
Therefore, the coordinates of point A in the new system after the transformation are:
\[ A'' = \left( \frac{-3}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]

### Summary of the Solution:
The coordinates of point A in the new system after the transformations are:
\[ \left( \frac{-3}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \]