What is the equation,in point-slope form,of the line parallel to the line y=3x+2 through point (-1,6)

Question

What is the equation,in point-slope form,of the line parallel to the line y=3x+2 through point (-1,6)

Answer 1

The equation, in point-slope form, of the line parallel to the line y=3x+2 through point (-1,6).

y – y₀ = m(x – x₀) is the equation, in point-slope form, of the line with slope m = 3 that passes through point (x₀,y₀).

A line parallel to the line y=3x+2 will have the same slope m = 3.

So the required equation of the line, in point-slope form is y – 6 = 3(x – (–1)) = y – 6 = 3(x + 1)

Answer 2

The original line y = 3x + 2 has a slope of 3.

So the line parallel to it must also have a slope of 3.

We don't yet know the y-intercept of the parallel line, so let's represent its equation leaving the y-intercept as a variable b.

y = 3x + b

To find the value of b, simply plug in the values of the point (-1,6). That is, x = -1 and y = 6. Then, solve for b.

(6) = 3(-1)+b

6+3 = b

9 = b

Great, we've found that b = 9. Let's plug it in:

y = 3x + b

y = 3x + 9

That's the answer: y = 3x + 9