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Step-by-Step Solution
Step 1: Write the equations of the lines
The line L is given by
$ \frac{x}{5} + \frac{y}{b} = 1. $
The line K, which is parallel to L, is given by
$ \frac{x}{c} + \frac{y}{3} = 1. $
Step 2: Determine the condition for parallelism
Two lines of the form
$ \frac{x}{a} + \frac{y}{m} = 1
\quad \text{and} \quad
\frac{x}{p} + \frac{y}{q} = 1 $
are parallel if the ratio of the coefficients of $x$ and $y$ are equal. Equivalently, their slopes are the same.
For L: rewriting
$ \frac{x}{5} + \frac{y}{b} = 1
\implies x \cdot \frac{1}{5} + y \cdot \frac{1}{b} = 1. $
The slope of this line is
$ -\frac{b}{5}. $
For K:
$ \frac{x}{c} + \frac{y}{3} = 1
\implies x \cdot \frac{1}{c} + y \cdot \frac{1}{3} = 1. $
The slope of this line is
$ -\frac{3}{c}. $
Since L and K are parallel, we have
$ \frac{b}{5} = \frac{3}{c} \quad \Longrightarrow \quad bc = 15. $
Step 3: Use the given point on line L to find b
The line L passes through the point $(13, 32)$. Substituting $x = 13$ and $y = 32$ into L, we get
$ \frac{13}{5} + \frac{32}{b} = 1. $
Rearrange to find $b$:
$ \frac{32}{b} = 1 - \frac{13}{5} = \frac{5}{5} - \frac{13}{5} = -\frac{8}{5}. $
Hence
$ \frac{32}{b} = -\frac{8}{5} \quad \Longrightarrow \quad b = -20. $
Step 4: Find c using bc = 15
We have $bc = 15$ and $b = -20$. Substituting,
$ -20 \cdot c = 15
\quad \Longrightarrow \quad c = -\frac{15}{20} = -\frac{3}{4}. $
Step 5: Rewrite line K in standard form
Using $c = -\frac{3}{4}$ and the equation
$ \frac{x}{c} + \frac{y}{3} = 1, $
we have
$ \frac{x}{-\frac{3}{4}} + \frac{y}{3} = 1
\quad \Longrightarrow \quad
\frac{-4}{3} x + \frac{1}{3}y = 1. $
Multiply through by 3 to get a simpler standard form:
$ -4x + y = 3 \quad \Longrightarrow \quad 4x - y + 3 = 0. $
Step 6: Find the distance between the two parallel lines
The line L can be rearranged in the form
$ \frac{x}{5} + \frac{y}{-20} = 1,
$
or
$ \frac{x}{5} - \frac{y}{20} = 1. $
This may be written as
$ \frac{1}{5}x - \frac{1}{20}y = 1. $
After clearing denominators, and putting in standard form, one can also express L similarly. However, a more straightforward approach is to use a point known to lie on L (for instance $(13,32)$) and measure perpendicular distance to K.
Using the standard distance formula from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$:
$ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. $
Let us choose a point on L, say $(13,32)$, which we already know. For line K in the form $4x - y + 3 = 0$, we have $A = 4$, $B = -1$, $C = 3$.
Substitute $(x_1, y_1) = (13, 32)$:
$ \text{Distance} = \frac{|4 \cdot 13 \;-\; 1 \cdot 32 \;+\; 3|}{\sqrt{4^2 + (-1)^2}}
= \frac{|52 - 32 + 3|}{\sqrt{16 + 1}}
= \frac{|23|}{\sqrt{17}}
= \frac{23}{\sqrt{17}}. $
Final Answer
The distance between lines L and K is
$ \frac{23}{\sqrt{17}}. $
