Question

The distance between the lines $3x+4y=9$ and $6x+8y+15=0$ is:

• A. $\frac{3}{10}$
• B. $\frac{33}{10}$
• C. $\frac{33}{5}$
• D. None of these

Answer 1

The correct option is B $\frac{33}{10}$

$3x+4y=9,6x+8y+15=0$

As can be seen from the slopes of the given lines, these two lines are parallel.

They can be written as $3x+4y=9$ and $3x+4y=\frac{-15}{2}$

Therefore, the perpendicular distance between these lines is obtained by $=\frac{C_1-C_2}{\sqrt{A^2+B^2}}=\frac{9+\frac{15}{2}}{\sqrt{9+16}}=\frac{\frac{33}{2}}{5}=\frac{33}{10}$