Find the positive distance, in units, between the parallel lines, with the given equations?

Question

Equations: 3x-y+9=0
6x-2y-4=0
Please show your process. Thanks.

Answer 1

y=3x+9 and y= 3x-2
The perpendicular to both through (0,0) is y=-1/3 x which cuts the firsT at
-1/3x =3x+9 IIIIIIII -10/3 x= 9 x =-27/10 y= 9/10 point A

The cut with the second is

-1/3 x =3x -2 IIIIIII -10/3 x= -2 x= 6/10 y= -2/10 point B

the distance between A and B is d=sqrt ((33/10)^2+(11/10)^2

= (sqrt 1210)/10 THIS IS THE DISTANCE BETWEEN THE PARALELS

Answer 2

First let us normalize the format of the two lines:
(1) y = 3x + 9
(2) y = 3x - 2

These are both lines with slope 3, so they are indeed parallel.

The distance between them means the length of a perpendicular line that intersects both of them.

There are infinitely many such perpendicular lines; they are all the parallel lines with slope -1/3. For example:
(3) y = -x/3

Now let us find the point on the first line, i.e. the intersection of (1) and (3)
y = 3 x + 9
y = -1/3 x or x = -3 y
y = -9 y + 9
10 y = 9, y = 0.9, x = -2.7
So point 1 is (-2.7, .9)

For point 2, we solve for the intersection of (2) and (3)
y = 3 x - 2
y = -1/3 x or x = -3 y
y = -9 y - 2
10 y = -2, y = -.2, x = .6
Point 2 is (.6, -.2)

Now solve for the distance (p1, p2)
dx = (.6 - -2.7) = 3.3
dy = (-.2 - .9) = -1.1
sqrt(dx^2 + dy^2) = sqrt(12.1)
The distance between the lines is approximately 3.48.

Answer 3

I dont know