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