Find the perpendicular distance between the lines
`9x+40y-20=0, 9x+40y+21=0`

Answer

To find the perpendicular distance between the lines given by the equations \(9x + 40y - 20 = 0\) and \(9x + 40y + 21 = 0\), we can use the formula for the distance between two parallel lines:

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

where the equations of the lines are in the form \(Ax + By + C = 0\). Here, \(A = 9\), \(B = 40\), \(C_1 = -20\) (for the first line), and \(C_2 = 21\) (for the second line).

### Step-by-Step Solution:

1. **Identify the coefficients:**
- From the first line \(9x + 40y - 20 = 0\), we have:
- \(A = 9\)
- \(B = 40\)
- \(C_1 = -20\)
- From the second line \(9x + 40y + 21 = 0\), we have:
- \(C_2 = 21\)

2. **Calculate the difference of the constants:**
\[ |c_1 - c_2| = |-20 - 21| = |-41| = 41 \]

3. **Calculate \(a^2 + b^2\):**
\[ a^2 + b^2 = 9^2 + 40^2 = 81 + 1600 = 1681 \]

4. **Calculate the square root:**
\[ \sqrt{a^2 + b^2} = \sqrt{1681} = 41 \]

5. **Substitute the values into the distance formula:**
\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{41}{41} = 1 \]

### Final Answer:
The perpendicular distance between the two lines is \(1\) unit.

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