Find the distance between the parallel lines `x+4sqrt(3)y+10=0` and `x+4sqrt(3)y-18=0`.

Answer

To find the distance between the parallel lines given by the equations \( x + 4\sqrt{3}y + 10 = 0 \) and \( x + 4\sqrt{3}y - 18 = 0 \), we can use the formula for the distance \( d \) between two parallel lines in the form \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \):

\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \]

### Step-by-Step Solution:

1. **Identify the coefficients from the equations**:
- From the first line \( x + 4\sqrt{3}y + 10 = 0 \):
- \( a = 1 \)
- \( b = 4\sqrt{3} \)
- \( c_1 = 10 \)
- From the second line \( x + 4\sqrt{3}y - 18 = 0 \):
- \( c_2 = -18 \)

2. **Calculate \( |c_1 - c_2| \)**:
\[ |c_1 - c_2| = |10 - (-18)| = |10 + 18| = |28| = 28 \]

3. **Calculate \( \sqrt{a^2 + b^2} \)**:
\[ a^2 = 1^2 = 1 \]
\[ b^2 = (4\sqrt{3})^2 = 16 \times 3 = 48 \]
\[ a^2 + b^2 = 1 + 48 = 49 \]
\[ \sqrt{a^2 + b^2} = \sqrt{49} = 7 \]

4. **Substitute into the distance formula**:
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{28}{7} = 4 \]

### Final Answer:
The distance between the two parallel lines is \( 4 \) units.

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