Question

The line $L_{1}:\frac{x}{5}+\frac{y}{b}=1$ passes through the point $\left(13 , 32\right)$ and is parallel to $L_{2}:\frac{x}{c}+\frac{y}{3}=1.$ Then, the distance between $L_{1}$ and $L_{2}$ is

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

Answer 1

Answer: (C)

Slope of the line $L_{1}=\frac{- b}{5}$ and slope of the line $L_{2}=\frac{- 3}{c}$ .
The line $L_{1}$ is parallel to the line $L_{2}$
$\Rightarrow \frac{b}{5}=\frac{3}{c}\Rightarrow bc=15$
As $\left(13 , 32\right)$ lies on $L_{1},$ so
$\frac{13}{5}+\frac{32}{b}=1$
$\Rightarrow \frac{32}{b}=\frac{- 8}{5}\Rightarrow b=-20$
So, $c=\frac{- 3}{4}$
$\therefore $ Equation of $L_{2}:y-4x=3$
Hence, distance between $L_{1}$ and $L_{2}$ $=\frac{\left|52 - 32 + 3\right|}{\sqrt{17}}=\frac{23}{\sqrt{17}}$ units