Question

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} }} \\\$

Answer 1

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.