Question

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

Answer 1

The correct option is D $\frac{23}{\sqrt{17}}$

$\begin{array}{c}\left(\frac{x}{5}\right)+\left(\frac{y}{b}\right)=1\left(1\right)pas\sin gthrough\left(13,22\right)\\ \Rightarrow \left(\frac{13}{5}\right)+\left(\frac{32}{b}\right)=1\\ \Rightarrow 13b+160=5b\\ \therefore b=-20\\ solineis-20x+5y=-100\left(1\right)\\ Secondline\\ \left(\frac{x}{3}\right)+\left(\frac{y}{3}\right)=1\\ 3x+cy=3c\left(2\right)\\ \left(1\right)and\left(2\right)areparallel\\ \left(\frac{3}{-20}\right)+5y=-9\times \left\{\frac{9}{4}\right\}\\ c=\left(\frac{-3}{4}\right)\\ Line3x-\left(\frac{3}{4}\right)y=-9\\ 12x-3y=-9\\ -20x+5y=-9\times \left(\frac{-5}{3}\right)\\ -20x+5y=\mathrm{15.......}\left(2\right)\\ Distancebetween\left(1\right)and\left(2\right)\\ =\frac{\left|-100+15\right|}{\sqrt{\left(400+25\right)}}\\ =\frac{115}{\sqrt{425}}\\ =\frac{115}{\overline{17}}\\ =\frac{23}{\sqrt{17}}\end{array}$

Hence, option $D$ is correct answer.