write an equation of the line that passes through (18,2) and is parallel to the line 3y-x= -12

Question

the assignment is writing equations of lines-parallel and perpendicular

Answer 1

If you graph this line first, you'll see that this is a trick question. Note that the original line already goes through the point (18,2). If you did not realize that, do the following.

As you know, parallel lines have equal slopes, so we need to determine the slope of the given line first. One way to do this is to convert it to slope intercept form.

3y-x=-12

3y = x-12 (add x to both sides)

y = 1/3 x -4 (divide both sides by 3)

No we know that the new line must have a slope of 1/3.

Start with the general form for all lines:

y = mx +b

Plug in the known slope

y = 1/3 x +b

Now, all we're short is the value of b. But we have a given point, which we can substitute in for x and y...

2 = 1/3 (18) +b

2 = 6 + b (simplifying)

-4 = b (subtracting 6 from both sides)

Now that we know the value of b, we know the new equation:

y = 1/3x -4 (Same as the original line)

Since parallel lines by definition have NO point of intersection, there is no solution.

Answer 2

Henry's answer is correct, but why not just use the way the original equation is written and plug in (x,y) = (18, 2)

3y - x = (3 * 2 - 18) = 6 - 18 = -12

3y - x = -12

It's prettier to write x - 3y = 12