Find distance between parallel lines 3x+4 and 3x-6

Question

Geometry question help please.

Answer 1

The distance between two parallel lines y₁=mx+b₁ and y₂=mx+b₂ is given by

d= |b₂-b₁|/sqrt(m²+1).

With m=3, b₁=4, and b₂=-6, you get

d= |-6-4|/sqrt(3²+1) = 10/√10 = √10.

Answer 2

The line perpendicular to both of this lines has a slope of -1/3 and the length of a segment of it, which is between two those lines, gives the distance between them. Let us consider the line y=-x/3. It is perpendicular to both lines. It intercepts y=3x+4 at the point, which x-coordinate satisfies the following equation:

 

-x/3=3x+4; or x=-6/5=-1.2; y-coordinate is y=-1.2/-3=0.4

 

Analogously, it intercepts the line 3x-6 at x=1.8

 

-x/3=3x-6 x=9/5=1.8

 

y-coordinate is 1.8/-3=-0.6

 

Distance between two lines, 3x+4 and 3x-6, is:

 

d=√[(1.8-(-1.2))²+(-0.6-0.4)²]=√(3²+(-1)²)=√10

 

Answer: √10