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 video transcript.

### Step 1: Find the Image of the Point with Respect to the Y-Axis
The original point is \( A(2, -1) \). To find the image of this point with respect to the Y-axis, we change the sign of the x-coordinate while keeping the y-coordinate the same.

\[ A' = (-2, -1) \]

### Step 2: Rotate the Coordinate Axes 45 Degrees Clockwise
When we rotate the coordinate axes 45 degrees clockwise, we need to find the new coordinates of the point \( A'(-2, -1) \) in the new system. The rotation of axes can be handled using the transformation formulas:

\[ x' = x \cos \theta + y \sin \theta \]
\[ y' = -x \sin \theta + y \cos \theta \]

Here, \( \theta = 45^\circ \), so \( \cos 45^\circ = \frac{1}{\sqrt{2}} \) and \( \sin 45^\circ = \frac{1}{\sqrt{2}} \).

### Step 3: Substitute the Coordinates into the Transformation Formulas
Substituting the coordinates of \( A'(-2, -1) \) into the transformation formulas:

\[ x' = (-2) \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{\sqrt{2}} = \frac{-2 - 1}{\sqrt{2}} = \frac{-3}{\sqrt{2}} \]

\[ y' = -(-2) \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{2 - 1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 4: Final Coordinates in the New System
Thus, the coordinates of point \( A' \) in the new coordinate system after the rotation 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:

\[ A''\left(\frac{-3}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]