Question

The line$L$given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to$L$and has the equation $\frac{x}{c}+\frac{y}{3}=1.$ Then, the distance between$L$and $K$ is

• A. $\frac{23}{\sqrt{15}}$
• B. $\sqrt{17}$
• C. $\frac{17}{\surd 15}$
• D. $\frac{23}{\surd 17}$

Answer 1

The correct option is D

$\frac{23}{\surd 17}$

Step 1: Given

1. $K$ and $L$ are two parallel lines.
2. Equation of $K$ is $\frac{x}{c}+\frac{y}{3}=1.$
3. Equation of $L$is$\frac{x}{5}+\frac{y}{b}=1$.
4. $L$passes through point $(13,32)$.

Step 2: Find the Distance between $K$and $L$.

The point $(13,32)$ lies on the line C

$$\begin{gathered} \Rightarrow \frac{x}{5}+\frac{y}{b}=1 \\ \Rightarrow \frac{13}{5}+\frac{32}{b}=1 \\ \Rightarrow b=-20 \end{gathered}$$

Now, the equation of $L$$=4x-y=20.......\left(1\right)$

The line $K$ and $L$are parallel

The slope of $K$ = The slope of $L$

$\Rightarrow 4=\frac{-3}{c}$

$\Rightarrow c=-\frac{3}{4}$

Putting c in the equation $K$

$$\begin{gathered} \Rightarrow \frac{x}{{\displaystyle \frac{-3}{4}}}+\frac{y}{3}=1 \\ \Rightarrow 4x-y=-3.......\left(2\right) \end{gathered}$$

Now, the distance between two parallel lines is $=\left|\frac{(c_2-c_1)}{\sqrt{a^2+b^2}}\right|$

where, $c_1$ and $c_2=$ constant in the equations $\left(1\right)$ and $\left(2\right)$ respectively.

$a$ and $b=$ coefficient of $x$ and $y$ in equations $\left(1\right)$ and $\left(2\right)$ respectively.

Distance $=\left|\frac{-3-\left(20\right)}{\sqrt{4^2+1^2}}\right|=\frac{23}{\sqrt{17}}$

Hence, the distance between $K$ and $L$is $\frac{23}{\surd 17}$.