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?

Answer 1

Given line L in $\frac{x}{5}+\frac{y}{b}=1$ passes through (13,32)

and line k=$\frac{x}{c}+\frac{y}{3}=1$ in paralled to

line L.

Now if (13,32) is a pount on line L then

$\frac{x}{5}+\frac{y}{5}$=1

$\frac{13}{5}+\frac{32}{5}$=1 $\Rightarrow \frac{32}{5}=1-\frac{13}{5}$

$\frac{b}{32}=\frac{-5}{8}$ $\Rightarrow [b=-20]$

line L = $\frac{x}{5}-\frac{y}{20}=1$

$4x-y=20------\left(1\right)$

if line k in parallel to line L then

3x+cy=3c----line k

if this is paralled then coeffcienth of x& y

as of line L& k are equal.

on comparing we get

$\frac{c}{3}=\frac{-1}{4}$$[c=\frac{-3}{4}]Hence$

Distance between line L& K in

$D=\left|\frac{20+\frac{3}{4}}{\sqrt{\left(1\right)2+\left(\frac{1}{4}\right)2}}\right|=\frac{83}{2\sqrt{17}}$