Question

Find the distance between the lines $2x+3y=10$ and $6x+9y=11$

Answer 1

The given lines are $2x+3y=10$ .......$\left(1\right)$

and $6x+9y=11$ .........$\left(2\right)$

Slope of line $\left(2\right)=\frac{-6}{9}=\frac{-2}{3}=$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

$3x-10=0\Rightarrow x=\frac{10}{3}$

$\therefore$Required distance between the given lines

$=$Perpendicular distance from $(\frac{10}{3},0)$ to the line $\left(2\right)$

$\frac{|6\times \frac{10}{3}+9\times 0-11|}{\sqrt{6^2+9^2}}=\frac{20-11}{\sqrt{36+81}}=\frac{9}{\sqrt{117}}=\frac{9\sqrt{117}}{117}=\frac{\sqrt{117}}{13}$ units.