Question

The distance between the lines x+y+2=0 and 3x+3y-7=0 is

Answer 1

Answer:

-2.846

Step-by-step explanation:

x+y+2=0 => (1)x+(1)y+2=0

3x+3y-7=0 => (3)x+(3)y-7=0

When the lines are given by

ax+bx+c1=0

ax+bx+c2=0

the distance between them is given by

d = c2−c1/√a2+b2

distance between the given lines will be

d = (-7) - (2)/√1² + 3²

= -7-2/√1+9

= -9/√10

= -2.846 Ans.

Hope it helps you..