Question

What is the distance between the straight lines $3x+4y=9$ and $6x+8y=15$?

• A. $\frac{3}{2}$
• B. $\frac{3}{10}$
• C. $6$
• D. $5$

Answer 1

The correct option is B $\frac{3}{10}$

Consider $3x+4y=9...\left(1\right)$

$\Rightarrow y=-\frac{3}{4}x+\frac{9}{4}$

Now consider $6x+8y=15...\left(2\right)$

$\Rightarrow 3x+4y=\frac{15}{2}$

$\Rightarrow y=-\frac{3}{4}x+\frac{15}{8}$

These two lines are parallel because slope between them is $-\frac{3}{4}$

We know that, the distance between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.

Here $c_1=9,c_2=\frac{15}{2}$

Hence $d=\frac{|9-\frac{15}{2}|}{\sqrt{3^2+4^2}}=\frac{\left|\frac{3}{2}\right|}{\sqrt{25}}=\frac{|3/2|}{5}=\frac{3}{10}$