Question

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32).$ If 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)

$\frac{23}{\sqrt{17}}$

Slope of line $L=-\frac{b}{5}$

Slope of line $K=-\frac{3}{c}$

Line $L$ is parallel to line $K$.

$bc=15$

$(13,32)$ is a point on $L.$

$\frac{13}{5}+\frac{32}{b}=1$

$\Rightarrow \frac{32}{b}=-\frac{8}{5}$

$b=-20$

Hence, equation of $L$ : $y-4x+20=0$

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

Equation of $K:y-4x=3$

Distance between $L$ and $K=\frac{|20+3|}{\sqrt{4^2+1^2}}=\frac{23}{\sqrt{17}}$