Question

Which of the following equations represents a line that is parallel to y = 3x +2

and passes through the point, (1,6)?

Answer 1

y=3x+3 is slope-intercept form

y-6=3(x-1) is point-slope form

I don't know the form your equation(s) are in. There are other forms.

Explanation:

The slope-intercept form a line is y=mx+b where m is the slope and b is the y-intercept.

So the slope of y=3x+2 is 3.

Parallel lines have the same slope. So the slope of the line we are looking for has a slope of 3.

Therefore, our equation his this form y=3x+b.

We need to now find b.

We know a point (x,y) on the line.

y=3x+b using (x,y)=(1,6).

6=3(1)+b

6=3+b

Subtract 3 on both sides:

3=b

So the equation is y=3x+3

Point-slope form is another form we can put our line into

y-y1=m(x-x1)

where m is the slope and (x1,y1) is a point on the line.

We have m=3 and a point (x1,y1) on the line is (1,6).

y-6=3(x-1)

Answer 2

`\huge{\boxed{y=3x+3}}`

It could also be
`\boxed{y-6=3(x-1)}`

We can use point-slope to find this. Parallel lines have the same slope, and the given line has a slope of 3 (
`m` in
`y=mx+b`). This means that the parallel line will also have a slope of 3.

Point-slope form is
`y-y_1=m(x-x_1)`, where
`m` is the slope and
`(x_1, y_1)` is a known point on the line.

Plug in the values.
`\boxed{y-6=3(x-1)}` (point-slope form)

Distribute.
`y-6=3x-3`

Add 6 to both sides.
`\boxed{y=3x+3}` (slope-intercept form)