Question

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

Answer 1

Answer:

1, 3, 5

Step-by-step explanation:

it’s correct on edge.

Answer 2

Answer:

First option.

Third option.

Fifth option.

Step-by-step 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:

$y- 1 = \frac{1}{3}(x+2)$

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 $y- 1 = \frac{1}{3}(x+2)$

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

Since $m\neq -3$, this line is not perpendicular to the line $y- 1 = \frac{1}{3}(x+2)$

3) $y = -3x-\frac{5}{3}$

Since $m=-3$, this line is perpendicular to the line $y- 1 = \frac{1}{3}(x+2)$

4) $\frac{1}{3}x - 2$

Since $m\neq -3$, this line is not perpendicular to the line $y- 1 = \frac{1}{3}(x+2)$

5) $3x + y = 7$

Solving for "y":

$y =-3x+ 7$

Since $m=-3$, this line is perpendicular to the line $y- 1 = \frac{1}{3}(x+2)$