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 $L_2$ $=\frac{|52-32+3|}{\sqrt{17}}=\frac{23}{\sqrt{17}}$ units