The line passing through `(-1,pi/2)` and perpendicular to `sqrt3 sin(theta) + 2 cos (theta) = 4/r` is

Options

A. `2 = sqrt(3)r cos theta - 2r sin theta`
B. `5 =- 2sqrt(3)r sin theta +4r cos theta`
C. `2 = sqrt(3)r cos theta +2r cos theta`
D. `5 = 2 sqrt(3)r sin theta +4r cos theta`

Explanation

To solve the problem, we need to find the equation of the line that passes through the point \((-1, \frac{\pi}{2})\) and is perpendicular to the given polar equation \( \sqrt{3} \sin(\theta) + 2 \cos(\theta) = \frac{4}{r} \).

### Step-by-step Solution:

1. **Convert the Polar Equation to Cartesian Form:**
The given polar equation is:
\[ \sqrt{3} \sin(\theta) + 2 \cos(\theta) = \frac{4}{r} \]
In polar coordinates, we have:
\[ x = r \cos(\theta) \quad \text{and} \quad y = r \sin(\theta) \]
Therefore, we can express \(r\) as:
\[ r = \sqrt{x^2 + y^2} \]
The equation can be rewritten as:
\[ \sqrt{3} \frac{y}{r} + 2 \frac{x}{r} = \frac{4}{r} \]
Multiplying through by \(r\) (assuming \(r \neq 0\)):
\[ \sqrt{3} y + 2 x = 4 \]

2. **Find the Slope of the Given Line:**
The equation \(\sqrt{3} y + 2 x = 4\) can be rearranged to find the slope:
\[ 2x + \sqrt{3}y = 4 \implies \sqrt{3}y = -2x + 4 \implies y = -\frac{2}{\sqrt{3}}x + \frac{4}{\sqrt{3}} \]
The slope \(m\) of this line is:
\[ m = -\frac{2}{\sqrt{3}} \]

3. **Find the Slope of the Perpendicular Line:**
The slope of a line perpendicular to another line is the negative reciprocal of the original slope. Therefore, the slope \(m'\) of the line we want is:
\[ m' = \frac{\sqrt{3}}{2} \]

4. **Use the Point-Slope Form to Find the Equation of the Line:**
We have a point \((-1, \frac{\pi}{2})\) and the slope \(m' = \frac{\sqrt{3}}{2}\). Using the point-slope form of the equation of a line:
\[ y - y_1 = m'(x - x_1) \]
Substituting the values:
\[ y - \frac{\pi}{2} = \frac{\sqrt{3}}{2}(x + 1) \]

5. **Rearranging to Standard Form:**
Distributing the slope:
\[ y - \frac{\pi}{2} = \frac{\sqrt{3}}{2}x + \frac{\sqrt{3}}{2} \]
Adding \(\frac{\pi}{2}\) to both sides:
\[ y = \frac{\sqrt{3}}{2}x + \frac{\sqrt{3}}{2} + \frac{\pi}{2} \]

### Final Equation:
The equation of the line that passes through the point \((-1, \frac{\pi}{2})\) and is perpendicular to the given polar equation is:
\[ y = \frac{\sqrt{3}}{2}x + \left(\frac{\sqrt{3}}{2} + \frac{\pi}{2}\right) \]