Question

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

• A. $3/2units$
• B. $3/10units$
• C. $3/4units$
• D. $2/7units$

Answer 1

Answer: (B)

$3/10units$

Let $(-3,0)$ be any point on line $3x+4y+9=0$
Distance of $(-3,0)$ and line $6x+8y+15=0$ is
$\frac{|6(-3)+8\left(0\right)+15|}{\sqrt{6^2+8^2}}$
$=\frac{3}{10}$units
Option B is correct.