Find the equation of a line parallel to the line `x cos alpha+ y sin alpha=p` and passing through the mid-point of the line segment joining the points `(1,5)` and `(3,-3)`.

Answer

To find the equation of a line parallel to the line given by \( x \cos \alpha + y \sin \alpha = p \) and passing through the midpoint of the line segment joining the points \( (1, 5) \) and \( (3, -3) \), we can follow these steps:

### Step 1: Find the Midpoint of the Given Points
The midpoint \( M \) of two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is calculated using the formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
For the points \( (1, 5) \) and \( (3, -3) \):
\[ M = \left( \frac{1 + 3}{2}, \frac{5 + (-3)}{2} \right) = \left( \frac{4}{2}, \frac{2}{2} \right) = (2, 1) \]

### Step 2: Identify the Slope of the Given Line
The equation of the line is given as:
\[ x \cos \alpha + y \sin \alpha = p \]
We can rearrange this into the slope-intercept form \( y = mx + c \):
\[ y \sin \alpha = p - x \cos \alpha \]
\[ y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{p}{\sin \alpha} \]
From this, we can see that the slope \( m \) of the given line is:
\[ m = -\cot \alpha \]

### Step 3: Write the Equation of the Parallel Line
Since the line we want to find is parallel to the given line, it will have the same slope \( m = -\cot \alpha \). We can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 1) \) and \( m = -\cot \alpha \):
\[ y - 1 = -\cot \alpha (x - 2) \]

### Step 4: Simplify the Equation
Distributing the slope:
\[ y - 1 = -\cot \alpha \cdot x + 2 \cot \alpha \]
Adding 1 to both sides:
\[ y = -\cot \alpha \cdot x + 2 \cot \alpha + 1 \]

### Step 5: Rearranging to Standard Form
To convert this to the standard form \( Ax + By = C \):
\[ \cot \alpha \cdot x + y = 2 \cot \alpha + 1 \]
Thus, the equation of the line is:
\[ x \cos \alpha + y \sin \alpha = 2 \cos \alpha + \sin \alpha \]

### Final Answer
The equation of the line parallel to \( x \cos \alpha + y \sin \alpha = p \) and passing through the midpoint \( (2, 1) \) is:
\[ x \cos \alpha + y \sin \alpha = 2 \cos \alpha + \sin \alpha \]

---