Question

Find the distance between the lines $3x+4y=9$ and $6x+8y=15$

Answer 1

Solution:-

$L_1:3x+4y=9$

Here, $a=3,b=4,c_1=-9$

$L_2:6x+8y=15\Rightarrow 3x+4y=\frac{15}{2}$

Here, $a=3,b=4,c_2=-\frac{15}{2}$

As we know that, distance between the two lines $L_1:ax+by+c_1=0$ and $L_2:ax+by+c_2=0$ is

$d=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$

$\therefore$ distance between the given lines, $\left(d\right)=\frac{|(-\frac{15}{2})-(-9)|}{\sqrt{3^2+4^2}}$

$\Rightarrow d=\frac{\left|\left(\frac{-15+18}{2}\right)\right|}{\sqrt{25}}$

$\Rightarrow d=\frac{\left(\frac{3}{2}\right)}{5}$

$\Rightarrow d=\frac{3}{10}$

Hence, the distance between the lines $3x+4y=9$ and $6x+8y=15$ is $\frac{3}{10}$ units.