Question

Find the distance between the lines 3x + y - 12 = 0 and 3x + y - 4 = 0

• A. $\frac{16}{\sqrt{10}}$
• B. $\frac{12}{\sqrt{10}}$
• C. $\frac{4}{\sqrt{10}}$
• D. $\frac{8}{\sqrt{10}}$
• E. Answer Required

Answer 1

Answer: (E)

Answer Required

The given lines are

$3x+y-12=0$

$3x+y-4=0$

The lines are parallel. The distance between the lines is

$r=\frac{|-12-(-4)|}{\sqrt{3^2+1^2}}$

$=>r=\frac{|-8|}{\sqrt{10}}$

$=>r=\frac{8}{\sqrt{10}}$