The line `P Q` whose equation is `x-y=2` cuts the x-axis at `P ,a n dQ` is (4,2). The line `P Q` is rotated about `P` through `45^0` in the anticlockwise direction. The equation of the line `P Q` in the new position is `y=-sqrt(2)` (b) `y=2` `x=2` (d) `x=-2`

Options

A. `y=-sqrt(2)`
B. y=2
C. x=2
D. x=-2

Explanation

To solve the problem step by step, we will follow the given information and perform the necessary calculations.

### Step 1: Identify the points and the line equation
The line \( PQ \) has the equation \( x - y = 2 \). We can rewrite this in slope-intercept form:
\[ y = x - 2 \]
This line intersects the x-axis when \( y = 0 \):
\[ 0 = x - 2 \implies x = 2 \]
Thus, the point \( P \) where the line intersects the x-axis is \( P(2, 0) \).

### Step 2: Identify point Q
We are given that point \( Q \) is \( (4, 2) \).

### Step 3: Find the slope of line PQ
To find the slope of line \( PQ \), we can use the coordinates of points \( P(2, 0) \) and \( Q(4, 2) \):
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{4 - 2} = \frac{2}{2} = 1 \]

### Step 4: Rotate the line about point P through 45 degrees
When we rotate the line \( PQ \) about point \( P(2, 0) \) through \( 45^\circ \) in the anticlockwise direction, we need to find the new slope of the line.

The original slope is \( 1 \). The angle \( \theta \) corresponding to this slope is:
\[ \tan^{-1}(1) = 45^\circ \]
After rotating \( 45^\circ \) anticlockwise, the new angle becomes:
\[ 45^\circ + 45^\circ = 90^\circ \]
The slope of a line at \( 90^\circ \) is undefined, which means the line is vertical.

### Step 5: Write the equation of the new line
Since the line is vertical and passes through point \( P(2, 0) \), the equation of the line is:
\[ x = 2 \]

### Conclusion
The equation of the line \( PQ \) in the new position after rotating \( 45^\circ \) about point \( P \) is:
\[ \boxed{x = 2} \]