The line `x+ 2y=4` is-translated parallel to itself by `3` units in the sense of increasing x and is then rotated by `30^@` in the clockwise direction about the point where the shifted line cuts the x-axis.Find the equation of the line in the new position

Options

A. `y=tan (theta-30^(@))(x-4-3sqrt(5))`
B. `y=tan (30^(@)-theta)(x-4-3sqrt(5))`
C. `y=tan (theta+30^(@))(x+4+3sqrt(5))`
D. `y=tan (theta-30^(@))(x+4+3sqrt(5))`

Answer

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

### Step 1: Understand the given line
The equation of the line is given as:
\[ x + 2y = 4 \]
We can rearrange this into slope-intercept form (y = mx + b):
\[ 2y = -x + 4 \]
\[ y = -\frac{1}{2}x + 2 \]
From this, we can identify:
- The slope (m) = -1/2
- The y-intercept = 2
- The x-intercept can be found by setting y = 0:
\[ x + 2(0) = 4 \Rightarrow x = 4 \]
Thus, the x-intercept is 4.

### Step 2: Translate the line
The line is translated parallel to itself by 3 units in the positive x-direction. Therefore, the new line will have the same slope but a different intercept. The new equation can be obtained by replacing \( x \) with \( x - 3 \):
\[ (x - 3) + 2y = 4 \]
\[ x + 2y = 7 \]

### Step 3: Find the point where the new line cuts the x-axis
To find the x-intercept of the new line \( x + 2y = 7 \), set \( y = 0 \):
\[ x + 2(0) = 7 \Rightarrow x = 7 \]
So, the new line cuts the x-axis at the point \( (7, 0) \).

### Step 4: Rotate the line
Next, we need to rotate the line \( x + 2y = 7 \) by \( 30^\circ \) clockwise about the point \( (7, 0) \).

The slope of the line before rotation is \( -\frac{1}{2} \). The angle \( \theta \) corresponding to this slope can be calculated as:
\[ \theta = \tan^{-1}(-\frac{1}{2}) \]

After a clockwise rotation of \( 30^\circ \), the new angle \( \theta' \) will be:
\[ \theta' = \theta - 30^\circ \]

The new slope \( m' \) can be calculated using the tangent of the new angle:
\[ m' = \tan(\theta - 30^\circ) \]

Using the tangent subtraction formula:
\[ \tan(\theta - 30^\circ) = \frac{\tan \theta - \tan 30^\circ}{1 + \tan \theta \tan 30^\circ} \]
Where \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \tan \theta = -\frac{1}{2} \):
\[ m' = \frac{-\frac{1}{2} - \frac{1}{\sqrt{3}}}{1 + (-\frac{1}{2})(\frac{1}{\sqrt{3}})} \]

### Step 5: Calculate the new slope
Calculating the numerator:
\[ -\frac{1}{2} - \frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{2\sqrt{3}} - \frac{2}{2\sqrt{3}} = -\frac{\sqrt{3} + 2}{2\sqrt{3}} \]

Calculating the denominator:
\[ 1 - \frac{1}{2\sqrt{3}} = \frac{2\sqrt{3} - 1}{2\sqrt{3}} \]

Thus,
\[ m' = \frac{-\frac{\sqrt{3} + 2}{2\sqrt{3}}}{\frac{2\sqrt{3} - 1}{2\sqrt{3}}} = \frac{-(\sqrt{3} + 2)}{2\sqrt{3} - 1} \]

### Step 6: Write the equation of the new line
Now, we can use the point-slope form of the line equation:
\[ y - y_1 = m'(x - x_1) \]
Where \( (x_1, y_1) = (7, 0) \):
\[ y - 0 = m'(x - 7) \]
Thus, the equation of the new line is:
\[ y = m'(x - 7) \]

### Final Step: Conclusion
The required equation of the line after translation and rotation is:
\[ y = m'(x - 7) \]