The line L given by $\dfrac{x}{5} + \dfrac{y}{b} = 1$passes through the point $(13,32)$. If line K is parallel to L and has the equation $\dfrac{x}{c} + \dfrac{y}{3} = 1$, then find the distance between L and K:
$
  A)\,\sqrt {17} \\
  B)\,\dfrac{{17}}{{\sqrt {15} }} \\
  C)\,\dfrac{{23}}{{\sqrt {17} }} \\
  D)\,\dfrac{{23}}{{\sqrt {15} }} \\
$

Hint:In order to solve this question , we need to substitute the given point \[(13,32)\] in line L . This will give us the value of b. Then by using the formula of slope of a line, we can get the value of c. At last calculate the distance between two lines to get your answer.

Complete step-by-step answer:
As stated in the question , line L passes through point \[(13,32)\]
Therefore, point $(13,32)$ lies on the line L
Hence it will satisfy the equation \[\dfrac{x}{5} + \dfrac{y}{b} = 1\], where $x = 13\,\& \,y = 32$ is replaced by $x\,\&\, y$
$
   \Rightarrow \dfrac{{13}}{5} + \dfrac{{32}}{b} = 1 \\
   \Rightarrow \dfrac{{32}}{b} = 1 - \dfrac{{13}}{5} = - \dfrac{8}{5} \\
   \Rightarrow \dfrac{{32}}{b} = - \dfrac{8}{5} \\
   \Rightarrow b = - 20 \\
$
So the equation of line L becomes $\dfrac{x}{5} - \dfrac{y}{{20}} = 1$
Multiplying the equation by $20$ we get
$4x - y = 1$
Now as stated in the question , line L is parallel to line K and we know the slope of parallel lines is equal.
Therefore, slope of line L is equal to slope of line K
Slope of line L $ = - \dfrac{{{\text{coefficient of y}}}}{{{\text{coefficient of x}}}} = \dfrac{1}{4}$
Slope of line K$ = \dfrac{{ - \dfrac{1}{3}}}{{\dfrac{1}{c}}} = - \dfrac{c}{3}$
$ \Rightarrow \dfrac{1}{4} = - \dfrac{c}{3} \Rightarrow c = - \dfrac{3}{4}$
Hence the equation of line K becomes $\dfrac{{4x}}{{ - 3}} + \dfrac{y}{3} = 1$.
Multiplying both sides by $ - 3$ we get,
$4x - y = - 1$
As you can see both lines L and K have the coefficients of $x\,\&\, y$ same which justifies that lines L & K are parallel.
Now we have to calculate distance between lines L and K
The formula for calculating distance between two parallel lines $ = \left| {\dfrac{{{c_1} - {c_2}}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
Where ${c_1}\,\&\, {c_2}$ are their intercepts and $a\,\&\, b$ are coefficients of $x\,\&\, y$
$ \Rightarrow {c_1} = 20\,\,,\,\,{c_2} = - 3\,\,,\,\,a = 4\,\,,\,\,b = - 1$
So distance between L and K
$
   = \left| {\dfrac{{20 + 3}}{{\sqrt {{4^2} + 1} }}} \right| \\
   = \dfrac{{23}}{{\sqrt {17} }}\,units \\
$

So, the correct answer is “Option C”.

Note:If two lines are parallel then the coefficients of x and y are equal.Students should remember the formula for calculating the slope of line and distance between two parallel lines for solving these types of questions.