Mathematics: 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

Which lines are perpendicular to the line y – 1

allieb12334

05.01.2022

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

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