solve below please

Question

to find the distance between parallel lines with equations x + 3y = 6 and x + 3y = -14, you first have to find a point on one line, find the equation through that point perpendicular to the first line and then find the point where that perpendicular intersects the second line. The equation of a line perpendicular to x + 3y = 6 is 3x – y = 0. Find the point where 3x – y = 0 intersects x + 3y = -14.

Answer 1

Start with one of the lines, you get to pick either (it doesn't matter). Then pick a value for "x" (any value is fine but it makes sense to pick something easy like "0" for instance. Plug that value of "x" into the equation you picked and solve for "y". Now you have a point (x, y) that is on the line you picked.

Example: I'll pick the equation x + 3y = 6 and I'll pick x = 0 as my value of "x". Plugging x = 0 into the equation, I can solve for "y":

x + 3y = 6

0 + 3y = 6

3y = 6

y = 2

The point on the line is (0, 2)

Now find the slope of the line that I originally picked. It's easiest to put the equation in slope-intercept form (y = mx + b)

x + 3y = 6

3y = -x + 6

y = (-1/3)x + 2

The slope is the coefficient in front of the "x" so it is -1/3.

To find the slope of the line PERPENDICULAR to this, take the negative reciprocal. The get the reciprocal when you flip the fraction over so the reciprocal of -1/3 is -3/1 or -3. Now, take the negative of it so the slope of the line PERPENDICULAR to your original line is 3. The equation of that line will be y = 3x + b but you don't know the value of "b" however, remember that you do have a point on the original line that you want the perpendicular line to go through, (0, 2). So plug that point into y = 3x + b:

2 = 3(0) + b

b = 2

So the equation of your perpendicular line is y = 3x + 2

Now, find out where this line intersects the OTHER original line (x + 3y = -14). To do so, substitute "3x + 2" into this equation in place of "y":

x + 3y = -14

x + 3(3x + 2) = -14

x + 9x + 6 = -14

10x = -20

x = -2

You can plug this into either equation to find the y-value:

Using y = 3x + 2, plug in x = -2:

y = 3(-2) + 2

y = -6 + 2

y = -4

So the point where your perpendicular line intersects "x + 3y = -14" is (-2, -4)

Now, use the distance formula to determine the distance between the 2 points: (0, 2) and (-2, -4):

Answer 2

find the distance between parallel lines with equations x + 3y = 6 and x + 3y = -14,

there is an easier method in the x/y plane for distance calculation from point to line.

if point is (x₀,y₀) and line has equation y=mx+b

then it can be proved with similar right triangles

or can be proved with system of two equations that

d=|mx₀+b-y₀|/√(m²+1)

point (0,2) for (x₀,y₀) from line x+3y=6

y=-(1/3)x-(14/3) is the equation of other line in y=mx+b form

using formula

d=|0-(14/3)-2|/√((-1/3)²+1)

d=(20/3)/√(10/9)=20/√10=2√10