The line L given by `x/5 + y/b = 1` passes through the point (13,32).the line K is parallel to L and has the equation `x/c+y/3=1` then the distance between L and K is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Find the value of \( b \) in line \( L \)
The equation of line \( L \) is given by:
\[ \frac{x}{5} + \frac{y}{b} = 1 \]
Since line \( L \) passes through the point \( (13, 32) \), we can substitute \( x = 13 \) and \( y = 32 \) into the equation:
\[ \frac{13}{5} + \frac{32}{b} = 1 \]

### Step 2: Solve for \( b \)
Rearranging the equation 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:
\[ 32 \cdot 5 = -8b \implies 160 = -8b \implies b = -\frac{160}{8} = -20 \]

### Step 3: Rewrite the equation of line \( L \)
Substituting \( b = -20 \) into the equation of line \( L \):
\[ \frac{x}{5} + \frac{y}{-20} = 1 \implies \frac{x}{5} - \frac{y}{20} = 1 \]
Multiplying through by 20 to eliminate the denominators:
\[ 4x - y = 20 \implies y = 4x - 20 \]

### Step 4: Find the slope of line \( L \)
From the equation \( y = 4x - 20 \), we see that the slope \( m_L \) of line \( L \) is:
\[ m_L = 4 \]

### Step 5: Find the equation of line \( K \)
The equation of line \( K \) is given by:
\[ \frac{x}{c} + \frac{y}{3} = 1 \]
Rearranging gives:
\[ \frac{y}{3} = 1 - \frac{x}{c} \implies y = -\frac{3}{c}x + 3 \]
The slope \( m_K \) of line \( K \) is:
\[ m_K = -\frac{3}{c} \]
Since lines \( L \) and \( K \) are parallel, their slopes must be equal:
\[ 4 = -\frac{3}{c} \implies c = -\frac{3}{4} \]

### Step 6: Rewrite the equation of line \( K \)
Substituting \( c = -\frac{3}{4} \) into the equation of line \( K \):
\[ \frac{x}{-\frac{3}{4}} + \frac{y}{3} = 1 \implies -\frac{4}{3}x + \frac{y}{3} = 1 \]
Multiplying through by 3 gives:
\[ -4x + y = 3 \implies y = 4x + 3 \]

### Step 7: Find the distance between lines \( L \) and \( K \)
The distance \( d \) between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
For line \( L \) (rewritten as \( 4x - y - 20 = 0 \)):
- \( A = 4 \), \( B = -1 \), \( C_1 = -20 \)

For line \( K \) (rewritten as \( 4x - y - 3 = 0 \)):
- \( C_2 = -3 \)

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

### Final Answer
The distance between lines \( L \) and \( K \) is:
\[ \sqrt{17} \]