Question

Find the distance between the lines $6x-8y=13$ and $3x-4y=12$

Answer 1

The given lines are $6x-8y=13$ .......$\left(1\right)$

and $3x-4y=12$ .........$\left(2\right)$

Slope of line $\left(2\right)=\frac{-3}{4}=\frac{-6}{8}=$Slope of line $\left(1\right)$

$\Rightarrow$ the given lines are parallel.

To find the distance between these lines,we choose a point on $\left(1\right)$

On putting, $y=0$ in $\left(1\right)$ we get

$6x=13\Rightarrow x=\frac{13}{6}$

$\therefore$Required distance between the given lines

$=$Perpendicular distance from $(\frac{13}{6},0)$ to the line $\left(2\right)$

$\frac{|3\times \frac{13}{6}-4\times 0-12|}{\sqrt{3^2+4^2}}$

$=\frac{|\frac{13}{2}-13|}{5}$

$=\frac{\left|\frac{13-26}{2}\right|}{5}$

$=\frac{13}{10}$ units.