Question

Find the distance between the lines $12x+16y=45$ and $3x+4y=2$

Answer 1

The given lines are $12x+16y=45$ .......$\left(1\right)$

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

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

$12x=45\Rightarrow x=\frac{45}{12}=\frac{15}{4}$

$\therefore$Required distance between the given lines

$=$Perpendicular distance from $(\frac{15}{4},0)$ to the line $\left(2\right)$

$\frac{|3\times \frac{15}{4}+4\times 0-2|}{\sqrt{3^2+4^2}}=\frac{\frac{45}{4}-2}{5}=\frac{\frac{45-8}{4}}{5}=\frac{37}{20}$ units.