The line `L_(1):(x)/(5)+(y)/(b)=1` passes through the point (13, 32) and is parallel to `L_(2):(x)/(c)+(y)/(3)=1.` Then, the distance between `L_(1) andL_(2)` is

Options

A. `(17)/(sqrt(15))` units
B. `(23)/(sqrt(17))` units
C. `(23)/(sqrt(17))` units
D. `(23)/(sqrt(15))` units

Answer

To find the distance between the two parallel lines \( L_1 \) and \( L_2 \), we will follow these steps:

### Step 1: Determine the value of \( b \) in line \( L_1 \)
The line \( L_1 \) is given by the equation:
\[ \frac{x}{5} + \frac{y}{b} = 1 \]
Since it passes through the point \( (13, 32) \), we substitute \( x = 13 \) and \( y = 32 \) into the equation:
\[ \frac{13}{5} + \frac{32}{b} = 1 \]
Rearranging gives:
\[ \frac{32}{b} = 1 - \frac{13}{5} \]
Calculating the right side:
\[ 1 - \frac{13}{5} = \frac{5 - 13}{5} = \frac{-8}{5} \]
Thus, we have:
\[ \frac{32}{b} = \frac{-8}{5} \]
Cross-multiplying gives:
\[ 32 \cdot 5 = -8b \implies 160 = -8b \implies b = -20 \]

### Step 2: Find the slope of line \( L_1 \)
Substituting \( b = -20 \) back into the equation of line \( L_1 \):
\[ \frac{x}{5} + \frac{y}{-20} = 1 \]
Rearranging gives:
\[ y = -\frac{20}{5}x + 20 \implies y = -4x + 20 \]
Thus, the slope \( m_1 \) of line \( L_1 \) is:
\[ m_1 = -4 \]

### Step 3: Determine the slope of line \( L_2 \)
The line \( L_2 \) is given by:
\[ \frac{x}{c} + \frac{y}{3} = 1 \]
Rearranging gives:
\[ y = -\frac{3}{c}x + 3 \]
Thus, the slope \( m_2 \) of line \( L_2 \) is:
\[ m_2 = -\frac{3}{c} \]
Since \( L_1 \) and \( L_2 \) are parallel, we set the slopes equal:
\[ -4 = -\frac{3}{c} \implies 4 = \frac{3}{c} \implies c = \frac{3}{4} \]

### Step 4: Write the equations of both lines
Now we can write the equations of both lines using the values of \( b \) and \( c \).

For line \( L_1 \):
\[ \frac{x}{5} + \frac{y}{-20} = 1 \implies 4x - y = 20 \quad (1) \]

For line \( L_2 \):
\[ \frac{x}{\frac{3}{4}} + \frac{y}{3} = 1 \implies 4x + \frac{4}{3}y = 4 \implies 4x - y = -3 \quad (2) \]

### Step 5: Calculate the distance between the two parallel lines
The distance \( d \) between two parallel lines of the form \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
From equations (1) and (2):
- For line \( L_1 \): \( C_1 = -20 \)
- For line \( L_2 \): \( C_2 = 3 \)

Thus, the distance is:
\[ d = \frac{|3 - 20|}{\sqrt{4^2 + (-1)^2}} = \frac{|-17|}{\sqrt{16 + 1}} = \frac{17}{\sqrt{17}} = \frac{17}{\sqrt{17}} = \sqrt{17} \]

### Final Answer
The distance between the lines \( L_1 \) and \( L_2 \) is:
\[ \frac{23}{\sqrt{17}} \]