Question

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $\left(13,32\right)$. 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. $\sqrt{17}$
• B. $\frac{17}{\sqrt{15}}$
• C. $\frac{23}{\sqrt{17}}$
• D. $\frac{23}{\sqrt{15}}$

Answer 1

Answer: (C)

Slope of line $L=-\frac{b}{5}$
Slope of line $K=-\frac{3}{c}$
Line L is parallel to line k.
$\Rightarrow \frac{b}{5}=\frac{3}{c}\Rightarrow bc=15$
$\left(13,32\right)$ is a point on L.
$\Rightarrow \frac{13}{5}+\frac{32}{b}=1\Rightarrow \frac{32}{b}=\frac{8}{5}$
$\Rightarrow b=-20\Rightarrow c=-\frac{3}{4}$
Equation of $K:y-4x=3$
Distance between L and $K=\frac{|52+32+3|}{\sqrt{17}}=\frac{23}{\sqrt{17}}$