Question

Find the distance between the given lines 4x - 3y = 12 and 4x - 3y = -8

• A. 3
• B. 4
• C. 5
• D. 6
• E. Answer Required

Answer 1

The correct option is B 4

The given lines are

$4x-3y=12$ $=>4x-3y-12=0$

$4x-3y=-8$ $=>4x-3y+8=0$

The lines are parallel. The distance between the lines is

$r=\frac{|8-(-12)|}{\sqrt{4^2+3^2}}$

$=>r=\frac{20}{5}$

$=>r=4$