Question

Which lines are perpendicular to the line y – 1 = One-third(x+2)? Check all that apply.

y + 2 = –3(x – 4)
y − 5 = 3(x + 11)
y = -3x – Five-thirds
y = One-thirdx – 2
3x + y = 7

Answer 1

1, 3, 5

Explanation:

it’s correct on edge.

Answer 2

First option.

Third option.

Fifth option.

Explanation:

The equation of the line in Slope-Intercept form is:

`y=mx+b`

Where "m" is the slope and "b" is the y-intercept.

The equation of the line in Point-slope form is:

`y - y_1 = m(x-x_1)`

Where "m" is the slope and
`(x_1,y_1)` is a point on the line.

By definition, the slopes of perpendicular line are negative reciprocals.

Then, given the line:

`y- 1 = (1)/(3)(x+2)`

We know that a line perpendicular to it, must have this slope:

`m=-3`

Let's check each option:

1)
`y + 2 = -3(x -4)`

Since
`m=-3`, this line is perpendicular to the line
`y- 1 = (1)/(3)(x+2)`

2)
`y - 5 = 3(x + 11)`

Since
`m\\eq -3`, this line is not perpendicular to the line
`y- 1 = (1)/(3)(x+2)`

3)
`y = -3x-(5)/(3)`

Since
`m=-3`, this line is perpendicular to the line
`y- 1 = (1)/(3)(x+2)`

4)
`(1)/(3)x - 2`

Since
`m\\eq -3`, this line is not perpendicular to the line
`y- 1 = (1)/(3)(x+2)`

5)
`3x + y = 7`

Solving for "y":

`y =-3x+ 7`

Since
`m=-3`, this line is perpendicular to the line
`y- 1 = (1)/(3)(x+2)`