Reduce the equation of the straight line `3x+4y+15=0` to normal form and find the perpendicular distance of the line from the origin.

Answer

To reduce the equation of the straight line \(3x + 4y + 15 = 0\) to normal form and find the perpendicular distance from the origin, we can follow these steps:

### Step 1: Write the equation in standard form
The equation is already in the standard form \(Ax + By + C = 0\), where \(A = 3\), \(B = 4\), and \(C = 15\).

### Step 2: Calculate the value of \( \sqrt{A^2 + B^2} \)
We need to find \( \sqrt{A^2 + B^2} \):
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Divide the entire equation by \( \sqrt{A^2 + B^2} \)
Now, we divide the entire equation by 5:
\[ \frac{3x}{5} + \frac{4y}{5} + \frac{15}{5} = 0 \]
This simplifies to:
\[ \frac{3}{5}x + \frac{4}{5}y + 3 = 0 \]

### Step 4: Rearrange the equation to the normal form
We can rearrange this to:
\[ \frac{3}{5}x + \frac{4}{5}y = -3 \]
This can be rewritten as:
\[ x \cos \alpha + y \sin \alpha = P \]
where \( \cos \alpha = \frac{3}{5} \) and \( \sin \alpha = \frac{4}{5} \).

### Step 5: Identify the perpendicular distance \( P \)
From the equation \(x \cos \alpha + y \sin \alpha = P\), we see that \(P = -3\). However, since distance cannot be negative, we take the absolute value:
\[ P = 3 \]

### Conclusion
The normal form of the line is:
\[ \frac{3}{5}x + \frac{4}{5}y = -3 \]
And the perpendicular distance from the origin to the line is:
\[ \text{Perpendicular distance} = 3 \]