Question

Find the distance between the lines $12x-16y=13$ and $3x-4y=12$

Answer 1

The given lines are $12x-16y=13$ .......$\left(1\right)$

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

Slope of line $\left(2\right)=\frac{-3}{4}=\frac{-12}{16}=$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

$12x=13\Rightarrow x=\frac{13}{12}$

$\therefore$Required distance between the given lines

$=$Perpendicular distance from $(\frac{13}{12},0)$ to the line $\left(2\right)$

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

$=\frac{|\frac{13}{4}-12|}{5}$

$=\frac{\left|\frac{13-48}{4}\right|}{5}$

$=\frac{35}{20}=\frac{7}{4}$ units.