Find the length of perpendicular from the point `(a cos alpha, a sin alpha)` to the line `x cos alpha+y sin alpha=p`.

Answer

To find the length of the perpendicular from the point \((a \cos \alpha, a \sin \alpha)\) to the line \(x \cos \alpha + y \sin \alpha = p\), we can follow these steps:

### Step 1: Identify the point and the line
The given point is:
\[ (x_1, y_1) = (a \cos \alpha, a \sin \alpha) \]
The equation of the line can be rewritten in the standard form:
\[ x \cos \alpha + y \sin \alpha - p = 0 \]
Here, we can identify:
- \(A = \cos \alpha\)
- \(B = \sin \alpha\)
- \(C = -p\)

### Step 2: Use the formula for the distance from a point to a line
The formula for the distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

### Step 3: Substitute the values into the formula
Substituting \(A\), \(B\), \(C\), \(x_1\), and \(y_1\) into the formula:
\[ d = \frac{|\cos \alpha (a \cos \alpha) + \sin \alpha (a \sin \alpha) - p|}{\sqrt{(\cos \alpha)^2 + (\sin \alpha)^2}} \]

### Step 4: Simplify the numerator
Calculating the numerator:
\[ \cos \alpha (a \cos \alpha) + \sin \alpha (a \sin \alpha) = a (\cos^2 \alpha + \sin^2 \alpha) \]
Using the Pythagorean identity \(\cos^2 \alpha + \sin^2 \alpha = 1\):
\[ = a \cdot 1 = a \]
Thus, the numerator becomes:
\[ |a - p| \]

### Step 5: Simplify the denominator
Calculating the denominator:
\[ \sqrt{(\cos \alpha)^2 + (\sin \alpha)^2} = \sqrt{1} = 1 \]

### Step 6: Final expression for the distance
Now substituting back into the distance formula:
\[ d = \frac{|a - p|}{1} = |a - p| \]

### Conclusion
The length of the perpendicular from the point \((a \cos \alpha, a \sin \alpha)\) to the line \(x \cos \alpha + y \sin \alpha = p\) is:
\[ \boxed{|a - p|} \]